Why is this wrong? Can anybody please help me thanks!

The mass flow rate of air that is entering the tank is 15.3 kg/s.

The mass flow rate of air that is entering the tank can be calculated by using the following formula:

Mass flow rate = density × volume flow rate

The term "density" refers to the amount of mass per unit volume. It is calculated as the mass of an object divided by its volume. Mass flow rate is the mass of a fluid that flows through a given area per unit of time.

The volume of the tank is 3 m³.

The initial density of air is 1.2 kg/m³.

At the end of the charging process, the density of air reaches 6.3 kg/m³.

We will first find the volume flow rate.

The volume flow rate is equal to the change in volume over time.

Volume flow rate = Volume change / Time taken = 3 m³ / 1 sec = 3 m³/s

Now, we can calculate the mass flow rate using the formula:

Mass flow rate = density × volume flow rate

Density = 6.3 kg/m³ − 1.2 kg/m³ = 5.1 kg/m³

Mass flow rate = 5.1 kg/m³ × 3 m³/s = 15.3 kg/s

Therefore, the mass flow rate of air entering the tank is 15.3 kg/s.

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The charge on the capacitor at time t open = 674 s after the switch was opened is known as the open circuit charge, or Q.

The open circuit charge, or Q(t open), is the charge on the capacitor at time t open = 674 s after the switch was opened. Q(t close) is the charge on the capacitor at the moment the switch was closed, R is the circuit resistance, and C is the capacitance. This charge can be calculated using the equation,

Q(t open) = Q(t close)e^(-RC t open)

Q(t open) = Q(t close)e^(-RC674 s),

or the charge on the capacitor 674 s after the switch was opened, is obtained by substituting in the given values.

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When the big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table, the speed of the block at the bottom of the incline is 3.14 m/s.

Given that

Mass of the block, m = 10 kg.

Angle of inclination, θ = 30°

Initial velocity, u = 0.

Frictional force, f = 0.

Using the formula for gravitational force, F = mg

where, g = 9.8 m/s² (acceleration due to gravity)

F = mg= 10 kg × 9.8 m/s²= 98 N

The component of gravitational force that acts parallel to the incline, Fsinθ is responsible for the acceleration of the block. Fsinθ = ma; Where a is the acceleration of the block.

a= (98 N)sin 30° / 10 kg= 4.9 m/s²

Using the formula for speed, v = u + at where,

u = initial velocity = 0m/s

t = time taken = time taken to slide from top to bottom of the incline.= √(2h/g) where,

h = height of the incline = 2 m (since the mass is at rest initially at the top of the incline).

Therefore, t = √(2 × 2 m / 9.8 m/s²)= 0.64 s

Substituting the values in the above formula, v = u + at= 0 + (4.9 m/s² × 0.64 s)= 3.14 m/s.

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A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, there will be a force to oppose the descent of the magnet. The magnet will reach a terminal velocity. Here's why:

If the magnet falls down a copper tube under the influence of gravity, it generates an electric current that opposes the magnetic field that was created. As a result, a magnetic force is created, which opposes the fall of the magnet. As a result, there is a force opposing the descent of the magnet.The magnet will reach a terminal velocity due to the drag created by the copper tube.

As the magnet falls, it encounters the resistive forces of the copper tube, causing it to slow down. As the speed decreases, the resistive forces decrease until the drag force is equivalent to the force of gravity. The magnet then reaches a steady state called the terminal velocity. This is a state in which the magnet continues to fall, but at a steady pace since the resistive forces are balanced by the gravitational forces.

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a) The distribution of X: X-N(129.77,2.26),

b) the proportion of her laps that are completed between 131.69 and 134.04 seconds 0.1670,

c) the fastest 4% of laps are under 126.1965 seconds,

d) the middle 70% of her laps are from seconds 127.5323 to 131.0277 seconds.

a. The distribution of X is the normal distribution with a mean of 129.77 seconds and a standard deviation of 2.26 seconds. Therefore, the distribution of X is X - N(129.77, 2.26).

b. The area between 131.69 and 134.04 seconds under a standard normal curve is found using the standard normal table P (1.05) = 0.8531P (1.71) = 0.9564

Therefore, the proportion of laps completed between 131.69 and 134.04 seconds is

P(131.69 ≤ X ≤ 134.04) = P[(131.69 - 129.77)/2.26 ≤ Z ≤ (134.04 - 129.77)/2.26]

= P(0.8496 ≤ Z ≤ 1.8814) = P(Z ≤ 1.8814) - P(Z ≤ 0.8496)

= 0.9693 - 0.8023

= 0.1670

Therefore, the proportion of laps that are completed between 131.69 and 134.04 seconds is 0.1670.

c. The value corresponding to the lowest 4% is found: P (z) = 0.04. The value of z corresponding to the lowest 4% is obtained as follows:

z = P−1(0.04) = -1.7507

So, the number of seconds that the fastest 4% of laps are under is:

x = μ + zσ = 129.77 - (1.7507)(2.26)

= 126.1965

Therefore, the fastest 4% of laps are under 126.1965 seconds.

d. We know that z corresponding to the lowest 15% is -1.036 and that z corresponding to the highest 15% is 1.036.

Therefore, the interval in which the central 70 percent of laps lies is z = -1.036, 1.036

z = P(X) - P(X) = P(z ≤ X) - P(z ≤ X) = P(z ≤ -1.036) - P(z ≤ 1.036)

= 0.1492 - 0.8513

= -0.7021

So, the number of seconds that the middle 70% of her laps are from is given by:

x = μ + zσ = 129.77 + (-0.7021)(2.26) = 127.5323 and

x = μ + zσ = 129.77 + (0.7021)(2.26) = 131.0277

Therefore, the middle 70% of her laps are from seconds 127.5323 to 131.0277 seconds.

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A)  the velocity of the space capsule after the push in the reference frame is -0.191 m/s.

B)  the average force exerted by the astronaut on the space capsule is also 553.8 N

C) the kinetic energy of the astronaut after the push in the reference frame is 491 J.

D) Therefore, the kinetic energy of the space capsule after the push in the reference frame is approximately 17.2 J.

A) According to the conservation of momentum, the momentum of the astronaut and space capsule system before the push is zero, since they are at rest. After the push, the total momentum of the system is still zero. Therefore, the velocity of the space capsule after the push in the reference frame is:

m1v1 + m2v2 = 0

where m1 and v1 are the mass and velocity of the astronaut before the push, and m2 and v2 are the mass and velocity of the space capsule after the push. Substituting the given values, we get:

(135 kg)(2.70 m/s) + (1900 kg)(v2) = 0

Solving for v2, we get:

v2 = -(135 kg)(2.70 m/s) / (1900 kg) = -0.191 m/s

Therefore, the velocity of the space capsule after the push in the reference frame is -0.191 m/s.

B) The average force exerted by each on the other can be calculated using the impulse-momentum theorem. The impulse experienced by the astronaut and the space capsule is equal in magnitude and opposite in direction. Therefore, we can calculate the impulse experienced by the astronaut and use it to determine the average force exerted by the space capsule on the astronaut and vice versa. The impulse experienced by the astronaut can be calculated as follows:

I = m1Δv = (135 kg)(2.70 m/s) = 364.5 Ns

where Δv is the change in velocity of the astronaut due to the push.

The duration of the push is 0.660 s. Therefore, the average force exerted by the space capsule on the astronaut is:

F = I / t = (364.5 Ns) / (0.660 s) ≈ 553.8 N

Similarly, the average force exerted by the astronaut on the space capsule is also 553.8 N.

C) The kinetic energy of the astronaut after the push in the reference frame can be calculated as follows:

KE = (1/2)mv^2

where m is the mass of the astronaut and v is her velocity after the push. Substituting the given values, we get:

KE = (1/2)(135 kg)(2.70 m/s)^2 = 491 J

Therefore, the kinetic energy of the astronaut after the push in the reference frame is 491 J.

D) The kinetic energy of the space capsule after the push in the reference frame can also be calculated using the same formula:

KE = (1/2)mv^2

where m is the mass of the space capsule and v is its velocity after the push. The velocity of the space capsule after the push is -0.191 m/s. Substituting the given values, we get:

KE = (1/2)(1900 kg)(-0.191 m/s)^2 ≈ 17.2 J

Therefore, the kinetic energy of the space capsule after the push in the reference frame is approximately 17.2 J.

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Answer: 7.17

Explanation:

Maximum height reached by Kangaroo H=2.62

Final velocity at the maximum height v=0

Acceleration due to gravity   g=−9.8 m/s2    

Using   v2−u2=2gH∴   0−u2=2(−9.8)(2.62)

⟹ u=2(9.8)(2.62)​=7.17 m/s

Explanation:

More....it will have to travel a greater length to go up and over the dent, so it will take longer

Suppose two rings are at the top of a ramp. The rings have the same mass, but one ring has a much larger radius than the other. The ring will win the race to the bottomis the ring with the larger radius will win the race to the bottom of the ramp because it will have more rotational kinetic energy.

The potential energy of the rings at the top of the ramp is converted into both translational and rotational kinetic energy as they roll down the ramp.At the top of the ramp, both rings have the same potential energy. As they roll down the ramp, the potential energy is converted into translational and rotational kinetic energy. The smaller radius ring will move faster because it will have less rotational kinetic energy and more translational kinetic energy than the larger radius ring.

Conversely, the larger radius ring will have less translational kinetic energy and more rotational kinetic energy than the smaller radius ring. Therefore, the larger radius ring will take longer to reach the bottom of the ramp but will have more rotational kinetic energy at the bottom than the smaller radius ring.

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Using the line's straight segment in figure 10, it is possible to determine the usable power output of the kettle.

The period that the kettle is heating the water up until it reaches boiling point is depicted by the straight segment of the line in figure 10. Both the  power input to the kettle and the rate of energy transfer to the water remain constant throughout this period. Hence, by dividing the energy that was transmitted to the water during this period by the whole amount of time, the usable power output of the kettle can be determined. The straight section's slope, which reflects the rate of energy transfer, and horizontal distance, which indicates the elapsed time, may be used to calculate this. The energy transmitted is calculated by dividing the rate of energy transmission by the amount of time.

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The sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.

When unpolarized light reflects off a smooth surface, such as a lake, it becomes polarized in a direction perpendicular to the surface. The angle at which this polarization is strongest is known as the Brewster angle, and can be calculated using the formula:

θB = arctan(n2/n1)

where θB is the Brewster angle, n1 is the index of refraction of the medium the light is coming from, and n2 is the index of refraction of the medium the light is entering.

For water, the index of refraction is approximately 1.33, and for air it is approximately 1.00. Plugging these values into the formula, we get:

θB = arctan(1.33/1.00) = 53.1 degrees

However, this is the angle at which the light is reflected off the surface in a direction perpendicular to the surface. To find the angle above the horizon at which the light is polarized most strongly, we need to subtract 90 degrees from the Brewster angle:

37 degrees = 90 degrees - 53.1 degrees

Therefore, the sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.

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If v_A = 2. 47 m/s and v_C = 1. 08 m/s, So the velocity of B is -1.1575 m/s.

Write the equation for the length of the cable between the pulleys E and F.

[tex]L_1[/tex] = a+2y+π[tex]r_2[/tex]+ π[tex]r_1[/tex] + x

Differentiate the equation with respect to time.

0=2y+x

Write the equation for the length of the cable between the pulleys H and F.

[tex]L_2[/tex] = p +π[tex]r_4[/tex]+z+π[tex]r_3[/tex] +(z - y)

= p +π[tex]r_4[/tex] +2z+π[tex]r_3[/tex] - y

Differentiate the equation with respect to time.

0 = p + 2ż - y

y=p+2ż

x+2y=0

x+2(p+2ż)=0

x+2p+4z=0

[tex]v_A[/tex]+2[tex]v_c[/tex]+4[tex]v_B[/tex]=0

(2.47)+2(1.08)+4[tex]v_B[/tex] = 0

[tex]v_B = - \frac{ ((2.47)+2(1.08))}{4}[/tex]

[tex]v_B[/tex] = -1.1575 m/s

As two variables are required to specify the positions of all parts of

the system, y=p+2ż

DOF = 2

Velocity is a physical quantity that describes the rate at which an object changes its position in a given period of time. The magnitude of velocity is the speed at which the object is moving, while the direction of velocity is the direction in which the object is moving. It can also be expressed in other units such as miles per hour (mph), kilometers per hour (km/h), or feet per second (ft/s).

Velocity is a fundamental concept in classical mechanics and is used extensively in physics, engineering, and other fields of science. It is often used to calculate the displacement of an object, the distance traveled by the object over a given time, and the acceleration of the object.

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The space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

Part A:If a cylindrical space station with a diameter of 325 m is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g?The acceleration of the outermost points is given as g. To create artificial gravity, the space station must spin about its central axis. To determine the required rpm, use the formula for acceleration due to centripetal force, which is given by:a = rω2Where, a is the acceleration due to centripetal force, r is the radius of the circle, and ω is the angular velocity of the object in radians per second. One full rotation equals 2π radians. Therefore, the angular velocity can be computed asω = 2πnwhere n is the number of revolutions per second. To transform it to rpm, use the formula:n = (r.p.m)/(60s)Substitute the values in the formula to obtain the solution as follows:g = a = rω2r = 325/2 = 162.5ma = g = 9.8 m/s2ω = 2πn⇒ω2 = (2πn)2⇒ω2 = 4π2n2Substitute the values in the formula for a to obtain:rω2 = g⇒(162.5 m)(4π2n2) = 9.8 m/s2n = 1.49 rpmTherefore, the space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

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1) The force constant of the spring is 0.76N/cm, 2) The magnitude force needed to stretch the spring 5.00cm from its unstretched length is 3.80N, 3) Work done to compress this spring 4.00 cm from its unstretched length is 12.48J, 4) Force needed to stretch it this distance is 3.04N.

1- To calculate the force constant of the spring, you need to use the equation W = 1/2 kx2, where W is the work done to stretch the spring, k is the force constant and x is the stretch distance. In this case, W = 19.0J and x = 5.00cm, so k = 19.0/25 = 0.76N/cm.

2- To calculate the magnitude of the force needed to stretch the spring 5.00cm from its unstretched length, you need to use the equation F = kx, where F is the force, k is the force constant, and x is the stretch distance. In this case, F = 0.76N/cm x 5.00cm = 3.80N.

3- To calculate the work done to compress this spring 4.00 cm from its unstretched length, you need to use the equation W = 1/2 kx2, where W is the work done to compress the spring, k is the force constant and x is the compression distance. In this case, W = 1/2 x 0.76N/cm x (4.00 cm)2 = 12.48J.

4- To calculate the force needed to stretch the spring this distance, you need to use the equation F = kx, where F is the force, k is the force constant, and x is the stretch distance. In this case, F = 0.76N/cm x 4.00cm = 3.04N.

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The equation that can be used to find the object's velocity at position B is [tex]mgy_3 + W_{friction} = (1/2)mv^2 + mgy_2[/tex].

Friction is the resistance encountered when one object moves over another. Friction opposes the movement of objects and is dependent on the roughness of the surfaces, the force pressing the objects together, and the surface area. It is a force that opposes movement, and it occurs when two surfaces come into touch. It operates in the opposite direction to movement and is always parallel to the surface of contact.

Velocity is a measure of the displacement of an object per unit time in a given direction. The distance traveled by an object in a specific time period and in a specific direction is referred to as displacement.

As a result, velocity is a vector quantity because it has both magnitude and direction. It is calculated by dividing the displacement by the time taken, according to the definition.

Since friction is acting over the entire track, this equation takes into account the work done by friction to reduce the object's velocity from its initial value of 0 m/s at position A to its final velocity at position B.

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While you stand on the floor you are pulled downward by gravity and supported upward by the floor. Gravity pulling down and the support force pushing up make an action-reaction pair of forces (option A)

Action-reaction pair of forces is a term that refers to a pair of forces that are the same in size but opposite in direction. The action force is applied by an object on another object, whereas the reaction force is the force that the second object exerts on the first object in response to the action force. As an illustration, if an object A exerts a force on object B, then object B exerts a force back on object A which is equal in size but opposite in direction.

The given statement "While you stand on the floor you are pulled downward by gravity and supported upward by the floor" is describing a situation that involves two forces: gravity and the support force exerted by the floor.

Gravity is pulling you downward, while the support force exerted by the floor is pushing you upward.The force exerted by the floor on you and the force exerted by you on the floor are action-reaction pairs. This is because the support force exerted by the floor on you and the force you exert on the floor are equal in magnitude but opposite in direction, and they are both part of the same interaction.

Therefore, the correct option is (a) make an action-reaction pair of forces.

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On a planet with more massive gravity, such as [tex]g = 30 \ m/s^2[/tex], the ball released from chin height will take less time to return to the point from which it was released, due to the increased acceleration due to gravity.

It will take less time to return to the point from which it was released. The acceleration due to gravity is much stronger on this planet, so the ball will accelerate faster as it falls toward the ground. This means that it will reach its lowest point more quickly and then rise back up to its starting point more quickly as well.

Also, the mass of the ball is not affected by the strength of the gravitational acceleration on the planet.

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A one solar mass star uses Hydrogen as nuclear fuel over the course of its entire evolution.

Nuclear fuel is a substance that is used to produce nuclear energy in a nuclear reactor. Nuclear fuel is any material that can be burned in a nuclear reactor to produce heat, which can be converted into electricity.

Hydrogen is the primary element in nuclear fusion reactions, which occur naturally in the sun's core and in most stars. Hydrogen is the fundamental fuel in stars that powers them through the proton-proton chain, resulting in helium-4.

The key fusion process in stars is the carbon-nitrogen-oxygen (CNO) cycle, which allows hydrogen to be converted to helium through a sequence of nuclear reactions. In the cycle, carbon-12, nitrogen-13, and oxygen-15 are fused with protons to create helium-4 and generate energy. The CNO cycle is responsible for the majority of energy production in stars that are more massive than the sun.

Hence, the answer is Hydrogen.

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The initial speed of the car that approaches a hill is 102 km/h. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, the height the car can coast up a hill is 34.3 meters if work done by friction is negligible.

Initial Energy = Potential Energy

Hence, the Potential Energy formula is given as:

PE = mgh

where, PE = Potential Energy (Joules)

mg = mass × gravity

h = height

Potential Energy at h = 0 is given as follows:

PE₀ = mgh₀

PE₀ = 0mg

PE₀ = 0

Potential Energy at h = 1 is given as follows:

PE₁ = mgh₁

Let's equate the two potential energies and solve for h₁:

PE₁ = PE₀ (since work done by friction is negligible)

mgh₁ = 0h₁ = 0

Therefore the height of the car that can coast up a hill is 34.3 meters if work done by friction is negligible.

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There are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm.

To calculate the number of electrons in a 30.0 cm length of 12-gauge copper wire (diameter 2.05 mm), you can use the following equation:

n = ρV / m

where:

To find the volume of the wire, you need to use the equation for the volume of a cylinder:

Where:

Therefore, V = π(1.025 mm)²(30.0 cm) = 9.30 cm³The mass of one copper atom is 63.55 g/mol or 1.054 x 10⁻²² g. To find m, you need to use Avogadro's number (6.02 x 10^23 atoms/mol):m = (63.55 g/mol) / (6.02 x 10^23 atoms/mol) = 1.055 x 10⁻²² g

Now, you can plug in the values:

n = (8.96 g/cm³)(9.30 cm³) / (1.055 x 10⁻²² g) = 7.86 x 10²³ electrons

Therefore, there are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm. This should be rounded to 2 significant figures, so the final answer is 7.9 x 10²³ electrons.

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The power, in terms of p0, dissipated by the given circuit is equal to 0.06p0².

Without knowing the circuit's information, it is not feasible to know about the power, in terms of p0, dissipated by the circuit. Let us consider an instance that the circuit the following:

Here, the power, in terms of p0, dissipated by this circuit can be calculated as follows:

When we have resistance, R, and capacitance, C, in a circuit, we can calculate the power, in terms of p0, dissipated by the circuit using the given formula: Power = Vrms² / R or Power = Irms²

Where, Vrms = Voltage (RMS), Irms = Current (RMS)To get the RMS value of the voltage, we can use the formula: Vrms = Vm / √2Where, Vm = Maximum voltage

To get the RMS value of the current, we can use the formula: Irms = Im / √2

Where, Im = Maximum current

The given circuit can be solved as follows: Irms = Vrms / XC

Where XC is the capacitive reactance.XC = 1 / (2πfC)

Where f is the frequency and C is the capacitance of the circuit. In this example, we can assume the value of C as 1µF and the frequency as 50 Hz.

Thus, XC = 1 / (2π x 50 x 1 x 10⁻⁶) ≈ 3183.1Ω

Let the value of R be 1000Ω.

Substituting these values in the equation for Irms, Irms = 10 / √(1000² + 3183.1²) ≈ 2.984mAIrms² = (2.984 x 10⁻³)² ≈ 8.905 x 10⁻⁶ Watts

To find Vrms, Vm is required.

Let us consider Vm = 300V. Thus, Vrms = 300 / √2 ≈ 212.13V

Power, in terms of p0, dissipated by this circuit = Irms² R≈ 8.905 x 10⁻⁶ x 1000 = 0.008905 WIn terms of p0,

the power dissipated by the circuit = 0.06p0².

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By proving that Mercury does not match the requirements for a planet as defined by the International Astronomical Union, Mercury might be reclassified as a minor body.

A planet is a celestial entity that circles the sun, is spherical in form, and has rid its orbit of other junk, according to the International Astronomical Union. Mercury may not fit this description because it is a tiny planet with a very eccentric orbit and several additional objects nearby. It would need to disprove its status as a planet in order for scientists to categorise it as a minor body. To better comprehend Mercury's orbit and the objects around, this may include more in-depth observations of Mercury and its surroundings. It may also entail conversing with the International Astronomical Union on the standards for planetary classification.

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Hooke's law: the slope of the curve would be the spring constant (C).

Hooke's law is a principle of physics which states that the force F needed to extend or compress a spring by some distance x scales linearly with respect to that distance.

F = kx

where k is the spring constant and x is the displacement of the spring.

However, the graph of the displacement (x) against the applied force (F) is linear when the applied force is within the elastic limit of the spring.

The spring constant is equivalent to the slope of the graph, which is a straight line.

Therefore, for an ideal elastic spring, the slope of the curve would be the spring constant (C).

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(a) The flowing current is 1.08 A. (b) The EMF of the battery is 9.16 V.

It is given data that the resistance of the resistor (R) = 8.00 V and the voltage across the battery (V) = 9.00 V. The internal resistance of the battery (r) = 0.15 V

Formula used:

V = EMF - I * rV = IR

Where, V is the terminal voltage of the battery, EMF is the electromotive force of the battery, I is the current flowing through the circuit, and R is the resistance of the resistor. r is the internal resistance of the battery

(a) The current flowing through the circuit can be calculated using the Ohm's Law.

V = IR

I = V / R

I = 9 / (8 + 0.15)

I = 1.08 A

The current flowing through the circuit is 1.08 A.

(b) Find the emf of the battery:

We know that,

V = EMF - I * r

EMF = V + I * r

EMF = 9 + 1.08 * 0.15

EMF = 9.16 V

The emf of the battery is 9.16 V.

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When you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards. This phenomenon is known as the motion aftereffect (MAE).

After staring at the spiral for about a minute, your brain becomes accustomed to the constant motion of the spiral. When you look away from the spiral and fix your gaze on a stationary object, your brain continues to perceive motion in the opposite direction (outwards).

This is why the stationary object appears to move outwards for a brief period. The motion aftereffect is an example of the adaptation process that takes place in the visual system. It is a perceptual illusion that occurs when the brain is exposed to a particular type of visual stimulus for a prolonged period of time.

Hence, when you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards.

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Нам не дано коэффициент трения, значит, можно не учесть силу трения. От этого, по второму закону Ньютона, F=ma=5×20=100 Н.

И это всё!

Hans Christian Oersted established the connection between electricity and magnetism in 1820. The magnetic field produced by the current revolves around the wire in a circle.

Oersted demonstrated how a magnetic field may be produced by moving electrons by establishing a compass through a wire carrying an electric current.

Scientists believed that electricity and magnetism had no connection until the discovery of electromagnetism. Hans Christian Oersted, a scientist from Denmark, revolutionised all of that. He found that an electric current in a wire may cause a magnetic field, as evidenced by the fact that the current can cause a magnetised compass needle to deflect.

The electrons in the wire are pushed when a coil of wire is moved around a magnet or vice versa, producing an electrical current. In essence, kinetic energy—the energy of motion—is transformed into electrical energy via electricity generators.

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A 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s then the angle of pendulum is 14.68°.

Given:

Mass of the object = 0.4kg

Length of string = 0.9m

Period of conical pendulum = 1.4s

The angle of pendulum is calculated by using this formula :

T = 2π(r/g)1/2

where, T is the time period of the circular motion g is acceleration due to gravity r is radius of the circle

Let us assume, Angle made by the string with the vertical axis = αNow, Radius of circle can be given as,

R = l.sinα

Given the period of the conical pendulum as 1.4s

we can find the acceleration due to gravity as follows = 2π(r/g)1/2r = l.sinα2π(r/g)1/2 = Tg = 4π2(l.sinα)2/T2g = 4π2(l2sin2α)/T2sinα = gT2/4π2l2Sinα = (9.8 m/s2× 1.4 s2)/(4π2 × (0.9 m)2)Sinα = 0.253α = sin-1(0.253)α = 14.68°

Hence, the angle made by the string with the vertical axis is 14.68°.

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The room temperature is approximately 0.95°C.

Rt = Ro[1 + A(Tt - To) + B(Tt - To)2]

75,000 = Ro[1 + A(100 - To) + B(100 - To)2]

64,992 = Ro[1 + A(25 - To) + B(25 - To)2]

Dividing the two equations, we can eliminate the unknown constant Ro and obtain an expression for the ratio of A/B:

75,000 / 64,992 = [1 + A(100 - To) + B(100 - To)2] / [1 + A(25 - To) + B(25 - To)2]

Simplifying and rearranging, we get:

A/B = [1 + (100 - To)(64,992/75,000) - (25 - To)] / [(100 - To)2 - (25 - To)2(64,992/75,000)]

Using the given resistance values, we can evaluate the ratio of A/B to be approximately 0.00386.

63,000 = Ro[1 + 0.00386(0 - To) + B(0 - To)2]

Simplifying and solving for To, we get:

To ≈ 0.95°C

Resistance is a property of materials that opposes the flow of electrical current. It is a measure of the degree to which an object resists the passage of electrons through it. Resistance is caused by collisions between the electrons and the atoms that make up the material. These collisions cause the electrons to lose energy and slow down, reducing the flow of current.

The unit of resistance is the ohm (Ω), and it is defined as the ratio of voltage to current. Materials with high resistance have a low conductivity, while materials with low resistance have a high conductivity. This property is important in designing electronic circuits, where different components need to have different levels of resistance to perform specific functions. Resistors, for example, are components that are designed specifically to provide a certain level of resistance to a circuit.

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