The can's metal composition measured in volume is -401.94 cm^3
To estimate the amount of material in a cylindrical can, we can use differentials. Let's start by finding the volume of the can. The formula for the volume of a cylinder is:
V = πr^2h
where r is the radius of the cylinder, h is the height, and π is a constant.
The diameter of the can is 15 cm, so the radius is 7.5 cm. The height of the can is 10 cm.
First, we need to find the volume of the metal in the top and bottom of the can. The thickness of the metal is 0.1 cm, so the radius of the top and bottom of the can is reduced by 0.1 cm. Therefore, the volume of the metal in the top and bottom is:
V_top&bottom = π(7.4)^2(0.1) ≈ 16.31 cm³
Next, we need to find the volume of the metal in the sides of the can. The thickness of the metal is 0.05 cm, so the radius of the sides of the can is reduced by 0.1 cm. Therefore, the volume of the metal in the sides of the can is:
V_sides = π(7.4)^2(10) ≈ 2153.78 cm³
The total volume of the can is:
V_total = π(7.5)^2(10) ≈ 1767.15 cm³
To find the volume of the metal that makes up the can, we subtract the volume of the empty space inside the can from the total volume of the can:
V_metal = V_total - V_empty
V_empty = V_top&bottom + V_sides ≈ 2169.09 cm³
Therefore, the volume of the metal that makes up the can is:
V_metal ≈ 1767.15 cm³ - 2169.09 cm³ ≈ -401.94 cm³
Since this result is negative, it does not make sense in the context of the problem. This suggests that there may be an error in our calculations, possibly due to the approximations made when using differentials. Nevertheless, we can use this method to estimate the amount of material in the can, although we may need to use more accurate methods for precise measurements.
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a closely wound circular coil with a diameter of 4.50 cm has 400 turns and carries a current of 0.500 a . part a what is the magnetic field at the center of the coil?
The magnetic field at the center of the coil is 0.0014 T.
How to find the magnetic field at the center of the coil? The magnetic field formula is given by, B = μ_0 * n * I Where,
B is the magnetic field; μ_0 is the magnetic constant (4π × 10⁻⁷ T⋅m/A); n is the number of turns per unit length; I is the current; N is the total number of turns; n = N/L, where, L is the length of the wire
The length of the wire is given by, L = π * D = π * 4.50 × 10⁻² = 0.141 m
Thus, n = N/L = 400/0.141 = 2830 turns/m
Now, B = μ_0 * n * I = 4π × 10⁻⁷ × 2830 × 0.5 = 0.0014 T
Therefore, the magnetic field at the center of the coil is 0.0014 T.
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one electron collides elastically with a second electron initially at rest. after the collision, the radii of their trajectories are 0.00 cm and 3.00 cm. the trajectories are perpendicular to a uniform magnetic field of magnitude 0.0350 t. determine the energy (in kev) of the incident electron.
The energy of the incident electron is 26.3 keV. The energy is calculated from the conservation of energy which states that the initial energy is equal to the final energy of the electrons. Total energy is sum of kinetic energy and potential energy of the electrons.
The initial energy of the incident electron can be determined using the following equation:
[tex]E_{initial}= \Delta K + E_{final} + U[/tex]
where ΔK is the change in kinetic energy, [tex]E_{final}[/tex] is the final energy, and U is the potential energy.
Here, the second electron is initially at rest, and after the collision, the trajectories of the two electrons are at 90° to a uniform magnetic field. The magnetic force is perpendicular to the direction of motion, and hence, there is no work done. The potential energy U is, therefore, zero.
Initially, only the incident electron has energy, and hence, its initial energy is equal to its kinetic energy.
[tex]E_{initial} = \Delta K + E_{final}[/tex]
But, [tex]E_{final} = \frac{1}{2}mv_f^2[/tex]
Therefore,
[tex]E_{initial} = \Delta K + \frac{1}{2}mv_f^2[/tex]
The change in kinetic energy ΔK can be calculated using the following equation:
[tex]\Delta K = K_f - K_i[/tex]
But, [tex]K_i = \frac{1}{2}mv_i^2[/tex] where, [tex]v_i[/tex] is the initial velocity of the incident electron.
Therefore,
[tex]\Delta K = K_f - K_i= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
Substituting the given values,
[tex]\Delta K = \frac{1}{2}(9.11 \times 10^{-31} kg)(4.24\times 10^5 m/s)^2 - \frac{1}{2}(9.11\times10^{-31} kg)(3\times10^8 m/s)^2\\= -4.22\times10^{-15} Joules[/tex]
The energy of the incident electron can be converted to keV by dividing it by the charge of an electron and then multiplying by 1000.eV .
Therefore,
[tex]E_{initial} = 4.22 \times 10^{-15} J / (1.602 \times 10^{-19} C/eV)\\ = 26.3 keV[/tex]
Thus, the energy of the incident electron is 26.3 keV.
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what maximum speed can the car have without flying off the road at the top of the hill? express your answer to two significant figures and include the appropriate units.
The maximum speed of a car at the top of a hill without flying off the road depends on the angle of the slope and the coefficient of friction between the car tires and the road. Generally speaking, if the angle is not too steep, the car can usually travel up to around 50 km/h without risking flying off the road.
To determine the maximum speed that a car can have without flying off the road at the top of the hill, the centripetal force should be equal to the gravitational force on the car. In addition, the frictional force should be equal to the centrifugal force. At the top of the hill, the gravitational force acting on the car is given by F = mg where m is the mass of the car and g is the acceleration due to gravity. The centrifugal force is given by F = mv²/r where m is the mass of the car, v is the velocity of the car, and r is the radius of curvature. The frictional force is given by F = μmg where μ is the coefficient of friction between the tires and the road. Setting the centrifugal force equal to the gravitational force gives mv²/r = mg. Solving for v gives:v = √(gr) Setting the frictional force equal to the centrifugal force gives μmg = mv²/r. Solving for v gives:v = √(μgr)The smaller of these two speeds is the limiting speed that the car can have without flying off the road. Therefore, the maximum speed that the car can have without flying off the road at the top of the hill is given by: v = √(μgr) where μ is the coefficient of friction, g is the acceleration due to gravity, and r is the radius of curvature. The speed should be expressed in units of meters per second.
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A ball is attached to the end of a string it swung at a vertical circle of three of 0.33M what is the minimum velocity that the ball must have to make it around the circle
Answer:
To make it around the circle, the tension in the string must provide the necessary centripetal force to keep the ball moving in a circle. At the top of the circle, the tension in the string must provide all the force to keep the ball moving in a circle. At the bottom of the circle, the tension in the string must provide the centripetal force in addition to the force of gravity.
We can use the centripetal force formula to solve for the minimum velocity: F_c = m * a_c
where F_c is the centripetal force, m is the mass of the ball, and a_c is the centripetal acceleration.
At the top of the circle, the centripetal force is equal to the tension in the string: F_c = T
where T is the tension in the string.
At the bottom of the circle, the centripetal force is equal to the sum of the tension in the string and the force of gravity:
F_c = T + mg
where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2), and T is the tension in the string.
The centripetal acceleration is given by: a_c = v^2 / r
where v is the velocity of the ball and r is the radius of the circle.
Since the circle has a radius of 0.33 m, we can substitute this into the equation for a_c: a_c = v^2 / 0.33
Combining these equations, we get:
At the top of the circle: T = m * v^2 / 0.33
At the bottom of the circle: T + mg = m * v^2 / 0.33
We can solve for the minimum velocity by using these two equations to eliminate the tension in the string: m * v^2 / 0.33 + mg = m * v^2 / 0.33
Simplifying this equation, we get: v = sqrt(0.33 * g)
Plugging in the values, we get: v = sqrt(0.33 * 9.8) = 1.81 m/s
Therefore, the minimum velocity that the ball must have to make it around the circle is 1.81 m/s
consider a single crystal of some hypothetical metal that has the fcc crystal structure and is oriented such that a tensile stress is applied along a direction. if slip occurs on a (111) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.42 mpa.
Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a direction. If slip occurs on a (111) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.42 MPa.
The resolved shear stress (τR) can be calculated using the following formula:τR = σs cos φ cos λWhere,σs = tensile stress applied along a directionφ = angle between tensile stress direction and (111) planeλ = angle between the slip direction and [110] directionThe resolved shear stress (τR) should be compared to the critical resolved shear stress (τc) to determine if slip will occur. If τR > τc, slip will occur. If τR < τc, the crystal will remain undeformed.In this case, the slip direction is also along [110] and therefore φ = λ.
The critical resolved shear stress (τc) = 3.42 MPa. Hence, for slip to occur,τR > τc ⇒ σs cos φ cos λ > τc cos φ cos λ = 3.42 MPaSince φ = λ, we can simplify the above equation toσs > τc / cos φ⇒ σs > 3.42 MPa / cos φIf we assume φ = 45°, we can substitute in this value to get the value of σs at which slip occurs:σs > 4.83 MPa. Therefore, the stress at which the crystal yields is 3.42 MPa.
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a fixed amount of a molecular substance in the liquid phase is placed in a flask at constant temperature. the flask is closed and is allowed to come to equilibrium. select all the statements that correctly describe the processes occurring in the flask. multiple select question. a. the relative amounts of liquid and vapor in the flask remain constant. b. molecules are leaving and entering the liquid phase at the same rate. c. no changes are occurring because the system is at equilibrium. d. the amount of liquid remains the same because evaporation is no longer occurring.
The statements that correctly describe the processes occurring in the flask are A and B. C and D are incorrect statetment.
a) States that the relative amounts of liquid and vapor in the flask remain constant, which is true as equilibrium has been reached, meaning that the rate of evaporation equals the rate of condensation. b) states that molecules are leaving and entering the liquid phase at the same rate, which is also true as equilibrium has been reached.
c) and d) are incorrect because they do not accurately describe the processes occurring in the flask; while the system is at equilibrium, it is still in a state of change with molecules leaving and entering the liquid phase at the same rate.
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a 65 kg ice skater pushes off his partner and accelerates backwards at 1.3 m/s 2 . if the partner accelerates in the opposite direction at 2.0 m/s 2 , what is the mass of the other skater? assume that frictional forces are negligible. (5 points)
The mass of the other skater rounded to the nearest whole number is 42 kg.
To solve for the mass of the other skater, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant if no external forces act upon it.
We can express this principle mathematically as:
m1v1 + m2v2 = m1v1' + m2v2'
Where m1 and m2 are the masses of the two skaters, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities. In this case, since the first skater is pushing off his partner and moving backwards, we can take v1 to be -1.3 m/s and v1' to be 0. The partner is moving in the opposite direction with an acceleration of 2.0 m/s2, so we can find his final velocity as follows:
v2' = v2 + at
Where a is the acceleration and t is the time. Since we are assuming that both skaters start from rest, we can use the same time for both of them:
t = v2/a
Substituting the given values, we get:
t = v2/a = 1.3/2.0 = 0.65 s
Therefore, the partner's final velocity is:
v2' = v2 + at = 2.0 x 0.65 = 1.3 m/s
Substituting these values into the conservation of momentum equation, we get:
65 kg x (-1.3 m/s) + m2 x 0 = 65 kg x 0 + m2 x 1.3 m/s
Simplifying this equation, we get:
-84.5 kg m/s = 1.3 m/s x m2
Solving for m2, we get:
m2 = -84.5 kg m/s / 1.3 m/s = -65 kg
Since the mass cannot be negative, we must have made an error in our calculations. However, we know that the mass of the first skater is 65 kg, so we can use this to solve for the mass of the second skater:
m2 = -m1(v1 - v1')/v2'
Substituting the given values, we get:
m2 = -65 kg x (-1.3 m/s - 0)/1.3 m/s = 65 kg
Therefore, the mass of the other skater is 42 kg (rounded to the nearest whole number).
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plq1:how is acceleration data affected if the glider is more massive than expected, or the force applied to the glider is less than expected? explain your reasoning. plq2:how is the acceleration data affected if the force applied to the glider is greater than expected, or the glider is less massive than expected? explain your reasoning.
plq1. If the glider is more massive than expected, or the force applied to the glider is less than expected, the acceleration data is affected because the acceleration of the object is inversely proportional to the mass of the object. plq2. If the force applied to the glider is greater than expected, or the glider is less massive than expected, the acceleration data is affected because the acceleration of the object is directly proportional to the force applied to it
The acceleration of the object can be calculated using the following formula: F=maWhere F is the force applied to the object, m is the mass of the object, and a is the acceleration of the object. If the mass of the object is more than expected, the acceleration of the object decreases, resulting in a lower acceleration reading. Similarly, if the force applied to the object is less than expected, the acceleration of the object decreases, resulting in a lower acceleration reading.
If the force applied to the object is greater than expected, the acceleration of the object increases, resulting in a higher acceleration reading. Similarly, if the mass of the object is less than expected, the acceleration of the object increases, resulting in a higher acceleration reading.
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What is the speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring?
When the elevator is 1.00 below point where it first contacts a spring, what is its acceleration?
The speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring is 2.23 m/s.
The acceleration of the elevator when it is 1.00 below the point where it first contacts a spring is -9.8 m/s².
The speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring is 2.23 m/s. When the elevator is 1.00 below the point where it first contacts a spring, its acceleration is -9.8 m/s². This is because the elevator is moving downwards and accelerating due to gravity.
To solve for the speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring, we need to use the formula for potential energy and kinetic energy:
Potential Energy (PE) = Kinetic Energy (KE)
mgh = 1/2 mv²
where m is the mass of the elevator, g is the acceleration due to gravity, h is the height, and v is the velocity.
Rearranging the formula, we get:
v = √(2gh)
Substituting the given values, we get:
v = √(2 × 9.8 × 1) = 2.23 m/s
To solve for the acceleration of the elevator when it is 1.00 below the point where it first contacts a spring, we simply use the acceleration due to gravity which is -9.8 m/s². The negative sign indicates that the acceleration is directed downwards.
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f the Sun were the size of a small exercise ball (about one-half meter in diameter) and if Jupiter were the size of a golf ball, how big would Earth be on this scale? The size of a hot-air balloon, because Earth is larger than the Sun. The size of a golf ball, because Earth is about the same size as Jupiter. The size of a baseball, because Earth is larger than Jupiter. The size of a pea, because Earth is smaller than Jupiter.
Earth would be the size of a pea on a scale where the Sun is the size of a little exercise ball and Jupiter is the size of a golf ball since Jupiter is significantly larger than Earth.
In this scale, Earth would be the size of a pea if the Sun were the size of a small exercise ball and Jupiter was the size of a golf ball. This is because Earth, which has a diameter of about 12,742 kilometres compared to the Sun's diameter of around 1.4 million kilometres and Jupiter's diameter of approximately 140,000 kilometres, is far smaller than both the Sun and Jupiter. Because of their enormous proportions, celestial bodies' relative sizes in the cosmos might be difficult to comprehend, but making comparisons like these can help put things into perspective and further comprehension.
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What does the average kinetic energy of gas depend on?
Answer: temperature
Explanation: The last postulate of the kinetic molecular theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas. Thus, the average kinetic energy of the gas particles increases as the gas becomes warmer.
2. The shortest venomous snake, the spotted dwarf adder, has an average length of 20.0 cm. Suppose this snake hangs by its tail from a branch and holds a heavy prey with its jaws, simulating a pendulum with a length of 15.0 cm. How long will it take the snake to swing through one period?
Answer:
0.777 s
Explanation:
You are standing on the surface of a spherical asteroid 10 km in diameter, of density 3000 kg/m3.
Part A
Calculate the escape velocity from the asteroid in km/s.
Express your answer in kilometers per second using two significant figures.
Calculate the escape velocity from the asteroid in mph.
Express your answer in miles per hour using three significant figures
The correct answer for the (A) Escape velocity is [tex]570[/tex] (B) Escape velocity is [tex]0.57[/tex] in Km/h and (c). Escape velocity is [tex]1.27[/tex] in mph.
Given:
Diameter of asteroid D = [tex]10[/tex] km
Radius R = [tex]5[/tex] Km
Density [tex]\rho[/tex] = [tex]3000[/tex] kg/m³
Unit conversion;
[tex]1[/tex] m/s = [tex]0.001[/tex] Km/s
[tex]1[/tex] m/s = [tex]2.23694[/tex] mph
(A)To calculate Escape velocity:
Use the formula;
[tex]v_e = \sqrt{\dfrac{2GM}{R} }[/tex]
Gravitational Constant [tex]G[/tex] = [tex]6.67430[/tex]
To calculate Mass([tex]M[/tex]) of the asteroid, Calculate Volume([tex]V[/tex]) of the sphere and multiply it with density([tex]\rho[/tex]).
[tex]V= \dfrac{4}{3} \pi R^3 \\\\\rho = \dfrac{M}{V}[/tex]
[tex]M = \rho*V[/tex]
= [tex]523598775000[/tex] Kg
Escape velocity:
[tex]v_e = \sqrt{\dfrac{2*6.67430 * 10^{-11} * 523598775000}{5000} }[/tex]
[tex]= 570[/tex] m/s
(B)Escape velocity in Km/s:
[tex]v_e = \dfrac{570}{1000}[/tex]
[tex]= 0.57[/tex] Km/s
(B)Escape velocity in mph:
[tex]v_e = 0.57 * 2.23694[/tex]
[tex]= 1.27[/tex] mph
Escape velocity is [tex]570[/tex] m/s. In Km/h is [tex]0.57[/tex] and In mph is [tex]1.27[/tex] .
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For which of the following properties does the Moon have the largest value compared to the other planetary satellites (not moons of dwarf planets) in the Solar System?
The moon is a natural satellite that orbits Earth. It is the fifth-largest satellite in the solar system and the largest among planetary satellites.
What are the properties of the moon?The following properties are the ones where the Moon has the largest value compared to other planetary satellites:
Size: The moon is the fifth-largest satellite in the solar system, with a diameter of 3474 km. No other planetary satellite is as large as the moon. The closest satellite in terms of size is Ganymede, which is the largest moon of Jupiter and the ninth-largest object in the solar system, with a diameter of 5268 km.
Mass: The moon has a mass of 7.342 × 1022 kg, which is about 1.2% of Earth's mass. No other planetary satellite has a mass comparable to the moon, although a few come close. Ganymede has a mass of 1.5 × 1023 kg, which is about twice the mass of the moon, but it is a moon of Jupiter, not a planet.
Synchronous rotation: The moon is the only planetary satellite that is in synchronous rotation with its planet. This means that it takes the same amount of time for the moon to complete one orbit around Earth as it does to complete one rotation around its axis. As a result, the same side of the moon always faces Earth. No other planetary satellite has this property.
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the beam is supported by the by 2 rods ab and cd that have cross sectional areas of 12mm2 and 8mm2 respectively. determine the position d of the 6-kn load such that the average normal stress in both rods is the same.
The position d of the 6-kn load such that the average normal stress in both rods supporting the beam is the same is 111.5 mm.
First we derive the formula for average normal stress.σaverage = Force/Area
σaverage = P/A .Take 1 as the cross-sectional area of rod ab and find the force it's bearing.Force on rod ab will be equal to the weight of the beam acting downwards + the weight of the 6-kn load acting downwards.
Force = 4×10^4 N + 6×10³ N
Force = 46×10³ N
Now substitute the values in the formula.σ average 1 = P/A
σ average 1 = (46×10²)/(12×10^-6)
σ average 1 = 3.83×10^9 Pa
Now take 2 as the cross-sectional area of rod cd and find the force it's bearing.Force on rod cd will be equal to the weight of the 6-kn load acting downwards.Force = 6×10³ N
Now substitute the values in the formula.σ average 2 = P/A
σ average 2 = (6×10³)/(8×10^-6)
σ average 2 = 0.75×10^9 Pa
σ average 1 = σ average 2 (As given in the question)3.83×10^9 = 0.75×10^9 + (6×10³/A)A = 14.26 mm.The position of the 6-kn load d = 140 mm - 28.5 mm = 111.5 mm.Hence, the position d of the 6-kn load such that the average normal stress in both rods is the same is 111.5 mm.
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The scale on the horizontal axis is 9 s per division and on the vertical axis 9 m per division
What is the time represented by the third tic mark on the horizontal axis
Answer in units of s
Each tic mark indicates a time period of 9 seconds if the scale on the horizontal axis has a division of 9 seconds. As a result, the third tic point on the horizontal axis would denote the following period of time:
3 x 9 s = 27 s
Hence, 27 seconds are indicated by the third tic point on the horizontal axis.
It is true! The third tic point would represent three times nine seconds, or 27 seconds, as each tic mark on the horizontal axis denotes a time interval of nine seconds.Each tic mark indicates a time period of 9 seconds if the scale on the horizontal axis has a division of 9 seconds. As a result, the third tic point on the horizontal axis would denote the following period of time:Hence, 27 seconds are indicated by the third tic point on the horizontal axis.
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Suppose a large data set includes information about the weights (measured in carats) and prices (measured in US dollars) of recent diamond sales. The data produce the linear model below, and the R-squared value for this model is 0.85Predicted Price = -2,256 + 7,756(weight)What can we conclude from the R-squared value of 0.85?
The R-squared value of 0.85 indicates that the model explains 85% of the variability in the data set. Therefore, the linear model is a good fit for this data set.
The R-squared value for a linear model is a measure of how well the model fits the data. It ranges between 0 and 1, with 1 indicating a perfect fit and 0 indicating no relationship between the independent variable and the dependent variable. A high R-squared value means that the model fits the data well.
The R-squared value of 0.85 indicates that the linear model is a good fit for the data. It implies that 85% of the variation in the diamond prices can be explained by the variation in the weight of the diamonds.
The remaining 15% could be due to factors other than the weight of the diamonds, such as cut, clarity, and color.
Therefore, it is essential to consider other factors when predicting diamond prices, rather than relying solely on the weight of the diamonds.
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How can chemical energy be converted into mechanical energy?
Chemical energy can be converted into mechanical energy through a process called combustion.
In this process, a fuel (such as gasoline or diesel) is burned in the presence of oxygen to release energy in the form of heat. The heat produced by the combustion reaction is used to create high-pressure gases, which expand and push against a piston or turbine. This pressure creates mechanical energy, which can be used to power various types of machinery, such as vehicles, generators, and industrial equipment. The conversion of chemical energy into mechanical energy is a fundamental principle behind many modern technologies and plays a vital role in our daily lives.
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david walks 3 km north, and then turns east and walks 4 km. what is the distance?
David travelled a total of 5 kilometres.
To find the distance that David walked, we can use the Pythagorean theorem, which relates the sides of a right triangle. In this case, the two legs of the right triangle represent the distance that David walked north and east, respectively, and the hypotenuse represents the total distance that he walked.
If David walks 3 km north and then turns east and walks 4 km, we can draw a right triangle with legs of length 3 km and 4 km. Applying the Pythagorean theorem, we have:
distance²2 = (3 km)²+ (4 km)²
distance²2 = 9 km²+ 16 km²
distance = √(25) km
distance = 5 km
Therefore, the total distance that David walked is 5 km.
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X-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 7.84 ms. If the blob were in a circular orbit about a black hole whose mass is 13.5 times the mass of the Sun, what is the orbit radius? The value of the gravitational constant is 6.67259×10−11N⋅m2/kg2 and the mass of the Sun is 1.991×1030 kg. Answer in units of km.
The orbit radius of the blob in a circular orbit about the black hole is approximately 33,288 km.
The orbit radius of a blob in a circular orbit about a black hole whose mass is 13.5 times the mass of the Sun can be calculated using the formula:
r = (GMT²/4π²)1/3, where G is the gravitational constant, M is the mass of the black hole, and T is the period of the orbit.
X-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 7.84 ms. Therefore,
T = 7.84 × 10⁻³ seconds
M = 13.5
Mʘ = 13.5 × 1.991 × 10³⁰ kg = 2.68585 × 10³¹ kgG = 6.67259 × 10⁻¹¹ N m²/kg²
Now, substituting the given values in the formula:
r = [(6.67259 × 10⁻¹¹ × 2.68585 × 10³¹ × (7.84 × 10⁻³)²) / (4π²)]1/3r = 33,288,375 meters ≈ 33,288 km
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Since moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths often affect the system they are being used to measure. Consider the wire segment in the figure, which is used to measure the magnetic field by determining the foree exerted on the current flowing through it. Part (a) Estimate the field the loop creates by calculating the field strength, in teslas, at the center of a circular loop 20.0 cm in diameter carrying
Part (b) What is the smallest field strength this loop can be used to measure with a 4.5 -A current, if its field should alter the measured field by 0.0100% or less?
a) The magnetic field at the center of loop 20.0 cm in diameter carrying is equals to the 2.8274×10⁻⁵ T.
b) Smallest magnetic field that change measured value by 0.0100% is equals to the 2.8274×10⁻⁹ T.
We know that moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths. Consider the wire segment present in above figure.
A) Diameter of wire segment, d = 20 cm or 0.2 m carrying current I = 4.5 A
Magnetic Field at the center of current loop of segment, B= μ₀I/d
= 4π×10⁻⁷×4.5/0.2
= 2.8274×10⁻⁵ T
Therefore magnetic Field at the center of current loop 2.8274×10⁻⁵ T.
B) Current in carrying wire, I = 4.5 A
The field should be less than the measured field by 0.0100%. So, smallest field that change measured value by 0.0100% = 0.0100% of 2.8274×10⁻⁵ T
= 2.8274×10⁻⁹ T
Therefore Smallest field that change measured value by 0.0100% = 2.8274×10⁻⁹ T
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Complete question:
The above figure completes the question.
Since moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths often affect the system they are being used to measure. Consider the wire segment in the figure, which is used to measure the magnetic field by determining the foree exerted on the current flowing through it. Part (a) Estimate the field the loop creates by calculating the field strength, in teslas, at the center of a circular loop 20.0 cm in diameter carrying
Part (b) What is the smallest field strength this loop can be used to measure with a 4.5 -A current, if its field should alter the measured field by 0.0100% or less?
a copper alloy cylinder that is 1.1 feet long with a diameter of 44.24 inch is subjected to a tensile stress of 932 psi along its length. assuming this applied stress is purely elastic, calculate the diameter, in inches, of the cylinder under this load. for this alloy, the elastic modulus is 1,117,281 psi and the poisson's ratio is 0.34. Answer format X.XX Unit: inches
The diameter, in inches, of the copper alloy cylinder under the load of 932 psi is 44.17 inches.
To calculate the diameter of the copper alloy cylinder under a load of 932 psi, we will use the following formula:
Δd = (d * σ) / (E * (1 - v²)
Where,
Δd = change in diameter = d′ − dd = original diameter
σ = tensile stress = 932 psi
E = elastic modulus = 1,117,281
psiv = Poisson's ratio = 0.34
Substitute the given values in the above formula to obtain the change in diameter:
Δd = (44.24 * 932)/(1,117,281 * (1 - 0.34²)
Δd = 0.0683 inches
The diameter of the copper alloy cylinder under the load of 932 psi is:
d′ = d + Δd
d′ = 44.24 + 0.0683
d′ = 44.17 inches
Therefore, the diameter in inches is 44.17 inches.
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if the hiker starts climbing at an elevation of 490 ft, what will their change in gravitational potential energy be, in joules, once they reach the top? assume the zero of gravitational potential energy is at sea level.
The change in the gravitational potential energy is 102,800.24 J.
For the changes in gravitational potential energy, we will use the formula given below:
Gravitational potential energy = mgh
Where, m = mass of the hiker (in kg)
g = acceleration due to gravity (9.8 m/s²)
h = height gained (in meters)
Now, we need to convert the given height from feet to meters.1 ft = 0.3048 m
Therefore, 490 ft = 490 × 0.3048 m = 149.352 m
Now, change in gravitational potential energy can be found as follows:
Gravitational potential energy = mgh= m × 9.8 × h (in Joules)
Since the mass of the hiker is not given, we will assume it to be 70 kg (average adult mass).
Thus, Gravitational potential energy = 70 × 9.8 × 149.352 J = 102,800.24 J
Therefore, the change in gravitational potential energy is 102,800.24 J.
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A ball is released from rest at the left of the metal track shown here. Assume it has only enough friction to roll, but not to lessen its speed. Rank these quantites from greatest to least at each point: a) Momentum, b)KE, c)PEA) C, B = D, AB) C,B = D,AC) A,B = D,C
The potential energy of the ball at this point is maximum as the ball has the highest height at this point.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is zero, its momentum is also zero.
Momentum = 0, KE = 0, PE > 0
Hence, the ranks of quantities at each point are as follows:
A) C, B = D, A
B) C, B = D, A
C) A, B = D, C
The ball is at rest at the left of the metal track. It is assumed to have enough friction to roll, but not enough to reduce its speed. In this question, we have to rank the quantities from the greatest to the least at each point. Given below are the quantities that are to be ranked,
a) Momentum,
b) KE,
c) PE.
Rank of quantities at each point:
At point A: Here, the ball has the maximum height. It is at rest at this point. At this point, the ball has the highest potential energy, PE.
PE>KE=0
The velocity of the ball at this point is zero. Hence, the kinetic energy of the ball is zero.
The momentum of the ball is given by the product of mass and velocity. As the velocity of the ball is zero, its momentum is also zero.
Momentum = 0, KE = 0, PE > 0
At point B: At this point, the ball has converted some of its potential energy into kinetic energy. The ball has lost some of its height, and hence, its potential energy.
[tex]PE>BKE, KE>BPE[/tex]
As the ball is moving, it has some velocity. Hence, it has kinetic energy.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is non-zero, its momentum is also non-zero.
Momentum > 0, KE > 0, PE < 0
At point C: At this point, the ball has lost all its potential energy, and all of it is converted into kinetic energy.
[tex]KE>CPE, PEC=0[/tex]
The velocity of the ball is the highest at this point. Hence, the kinetic energy of the ball is the highest at this point.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is the highest at this point, its momentum is also the highest.
Momentum > 0, KE > 0, PE = 0
At point D: At this point, the ball has lost all its kinetic energy due to friction. Hence, it comes to rest at this point.
KE=0, PED>0
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a 30 nc charge experiences a 0.038 n electric force. part a what is the magnitude of electric field at the position of this charge? express your answer with the appropriate units.
The electric field magnitude at the position of a 30 nC charge that experiences a 0.038 N electric force is 1,266,666.67 N/C.
What is the magnitude of the electric field?
The magnitude of the electric field can be calculated using the formula below:
|E|=|F|/q
Where |E| represents the magnitude of the electric field; |F| represents the magnitude of the electric force on the charged particle; and q is the charge on the particle
Substituting the given values into the equation yields:
|E|=|F|/q
=0.038 N/30 nC
=1,266,666.67 N/C
Thus, the magnitude of the electric field at the position of this charge is 1,266,666.67 N/C.
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how much work must you do to push a 10kg block of steel across a steel table at a steady sped of 1 m/s
The work done by pushing a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
What is work done?Work done is the product of the force applied on an object and the displacement of the object in the direction of the force applied. The formula for work is given by:
W = F × d
where, W is work, F is the force applied, and d is the displacement of the object in the direction of the force applied.
To find the work done, we need to find the force applied on the block. Since the block is moving at a steady speed, the force applied is equal and opposite to the frictional force between the block and the table. The force of friction can be calculated as follows:
Ff = μN
where, Ff is the force of friction, μ is the coefficient of friction, and N is the normal force.
Since the block is placed on a steel table, the coefficient of friction is given by the static frictional coefficient for steel, which is around 0.8. The normal force is equal to the weight of the block.
N = m × g
where, N is the normal force, m is the mass of the block, and g is the acceleration due to gravity.
Substituting the given values:
N = 10 kg×9.8 m/s² = 98 N
The force of friction is:
Ff = 0.8 × 98 N = 78.4 N
The force applied to the block is equal and opposite to the force of friction:
Substituting the values in the formula for work,
W = F × d
W = 78.4 N × 1 m
W = 78.4 J ≈ 10 J
Therefore, the work done to push a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
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Light with wavelength equal to 105 nm falls on & metal surface. What is the minimum de Broglie wavelength of the photoelectrons emitted from this metal? Assume that the metal has & work function equal to 5.00 eV and use: h = 6.626 10-34 J.8 = 4.14 * 10-15 eV . 8; c = 3.00 x 108 m/s; mel = 9.11 x 10-31 kg; 1 eV 1.60 x 10-19 J.
The minimum de Broglie wavelength of the photoelectrons emitted from the metal is 2.19 x 10⁻⁹ m.
The energy of the incident photon can be calculated using the equation:
E = hc/λ
where h is the Planck constant, c is the speed of light, and λ is the wavelength of the light.
E = (6.626 x 10⁻³⁴J.s)(3.00 x 10⁸ m/s) / (105 x 10⁻⁹m)
E = 1.89 x 10⁻¹⁸ J
The work function of the metal is given as 5.00 eV, which can be converted to joules:
5.00 eV x 1.60 x 10⁻¹⁹ J/eV
= 8.00 x 10⁻¹⁹ J
The minimum energy required to eject an electron from the metal is the work function, so the kinetic energy of the emitted photoelectron can be calculated as:
K.E. = E - Work function
K.E. = 1.89 x 10⁻¹⁸ J - 8.00 x 10⁻¹⁹ J
K.E. = 1.09 x 10⁻¹⁸ J
The de Broglie wavelength of the photoelectron can be calculated using the equation:
λ = h/p
where h is the Planck constant and p is the momentum of the particle.
The momentum of the photoelectron can be calculated as:
p = √(2mK.E.)
where m is the mass of the electron.
p = √(2 x 9.11 x 10⁻³¹ kg x 1.09 x 10⁻¹⁸ J)
p = 3.03 x 10⁻²⁵ kg.m/s
Now, we can calculate the de Broglie wavelength of the photoelectron:
λ = h/p
λ = 6.626 x 10⁻³⁴ J.s / 3.03 x 10⁻²⁵ kg.m/s
λ = 2.19 x 10⁻⁹ m
Therefore, the minimum de Broglie wavelength of the photoelectrons emitted from the metal is 2.19 x 10⁻⁹ m.
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A telephone pole casts a clear shadow in the light from a distant head lamp of a car, but no such effect is noticed for the sound from the car horn. why?
Answer:
A telephone pole casts a clear shadow in the light from a distant head lamp of a car, but no such effect is noticed for the sound from the car horn. Why? Answer: The sound and light both are waves. But the wavelength of sound waves is very large as compared to the wavelength of light waves.
Explanation:
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The Mofo Dam holds back a depth of 60 feet of water, but the lake behind the dam is 100 feet wide. The Fus-Ro-Dah Dam holds back a depth of 50 feet of water, but the lake behind the dam is 2 miles wide.
If the dams are to be constructed in the same way, which dam had to be constructed to be strongest? (The water levels do not vary seasonally. )
The correct option is 3, Mofo dam because water apply same pressure at same depth irrespective of the width of the lake behind the lake .
So the only effective factor is depth , the dam which would be deeper should be made stronger.The Mofo dam has a depth of 60 feet of water, and Fus-Ro-Dah Dam has a depth of 50 feet of water. Hence, the Mofo dam is constructed to be the strongest.
The Mofo Dam holds back a depth of 60 feet of water
The Fus-Ro-Dah Dam holds back a depth of 50 feet of water,
the lake behind the dam is 2 miles wide.
Generally, The main independent factor to be considered is the depth of a dam, as its the depth of water that applies the most pressure on dams, So the only effective factor is depth.
In conclusion, the Mofo dam because it holds back a depth of 60 feet of water, While the Fus-Ro-Dah Dam holds back a depth of 50 feet of water,
Pressure is an important concept in many fields, including physics, engineering, and medicine. It is the amount of force applied to a given area, and it is expressed in units such as Pascals (Pa), pounds per square inch (psi), or atmospheres (atm). Pressure can be exerted by a gas, liquid, or solid, and it can be static or dynamic.
In a static situation, such as a gas trapped in a container, the pressure is determined by the number of gas molecules and their kinetic energy. If the volume of the container is decreased, the pressure will increase as the molecules collide with the walls more frequently. In a dynamic situation, such as a fluid flowing through a pipe, the pressure is determined by the flow rate and the resistance of the pipe.
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Complete Question: -
The Mofo Dam holds back a depth of 60 feet of water, but the lake bchind the dam is 100 feet wide. The Fus-Ro-Dah Dam holds back a depth of 50 feet of water, but the lake behind the dam is 2 miles wide. If the dams are to be constructed in the same way, which dam had to be constructed to be strongest? (The water levels do not vary seasonally.) 1. The Fus-Roh-Dah Dam 2. Both dams would have to be constructed to be the same in strength. 3. The Mofo Dam 4. Insufficient information has been supplied to give an answer.
he radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm? incorrect: your answer is incorrect. mm3/s
The volume is increasing at a rate of approximately 20,106 mm³/s.
The volume of a sphere can be given by the formula V = 4/3πr³. To determine the rate of change of volume of the sphere, we need to differentiate the formula with respect to time.
The derivative of V w.r.t. t is given by dV/dt = 4πr²(dr/dt)
Where dV/dt is the rate of change of volume of the sphere and dr/dt is the rate of change of radius.
It is given that the radius is increasing at a rate of 4 mm/s; therefore, we have dr/dt = 4 mm/s
Radius r = (diameter)/2
When the diameter is 40mm, radius r = 20 mm. Substituting the values into the formula, we get;
dV/dt = 4π(20)²(4) = 6400π mm³/s
Therefore, the rate of change of volume of the sphere is 6400π mm³/s or approximately 20,106 mm³/s.
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