Answer:
They were deposited horizontally and then shifted by a geologic event.
Explanation:
Problem 1. In this problem, you need to determine the additive inverse1of each given vector in the appropriate vector space. (a)[ 23]inR 2. (b)−1+3x−8x 2inP 2. (c)[ 12−20]inM 2×2.
The additive inverse of each given vector in the appropriate vector space are
(a) The additive inverse of [2, 3] in [tex]R_2[/tex] is [-2, -3].
(b) The additive inverse of [tex]-1 + 3x - 8x^2[/tex] in P2 is [tex]1 - 3x + 8x^2[/tex].
(c) The additive inverse of [1, 2; - 2, 0] in [tex]M_{2\times2[/tex] is [-1, -2; 2, 0].
The additive inverse of a vector [tex]\mathbf{v}[/tex] in a vector space is the vector [tex]-\mathbf{v}[/tex] that, when added to [tex]\mathbf{v}[/tex], gives the zero vector.
(a) The additive inverse of the vector [tex][2, 3] \in \mathbb{R}^2[/tex] is [tex][-2, -3][/tex] since [tex][2, 3] + [-2, -3] = [0, 0][/tex].
(b) The vector space [tex]P_2[/tex] consists of all polynomials of degree at most [tex]2[/tex]. The vector [tex]-1 + 3x - 8x^2 \in P_2[/tex] has additive inverse [tex]1 - 3x + 8x^2[/tex], since [tex](-1 + 3x - 8x^2) + (1 - 3x + 8x^2) = 0[/tex].
(c) The vector space [tex]M_{2 \times 2}[/tex] consists of all [tex]2 \times 2[/tex] matrices. The matrix [tex][1, 2; -2, 0] \in M_{2 \times 2}[/tex] has additive inverse [tex]$[-1, -2; 2, 0]$[/tex], since [tex]\begin{bmatrix} 1 & 2 \ -2 & 0 \end{bmatrix} + \begin{bmatrix} -1 & -2 \ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}[/tex].
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The colors on an oil slick are caused by reflection and (explain why)
a. Diffraction
b. Interference
c. Refraction
d. Polarization
e. Ionization
"The colours on an oil slick are caused by reflection and interference." Correct option is B.
Different bands of the oil slick create different colours as the oil film progressively thins from the centre to the edges.
Interference is what gives an oil slick drifting on water or a soap bubble in the sun their vibrant colours. The colours that interact most positively are the ones that are most vibrant. Thin film interference is the name given to the phenomenon because it occurs when light reflected from various thin film surfaces interferes with one another.
The most crucial interfering principle is the superposition principle.
This hair colour procedure primarily uses jewel tones and rainbow colours, including burgundy, royal blue, deep purple, green, and deep red. Alternating the colours that give your hair an oil spill appearance is the best method to make your skin tone and hair look good together. Best choice is B.
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Assume that a drop of mercury is an isolated sphere. What is the capacitance in picofarads of a drop that results when two drops each of radius R = 5.61 mm merge?
The formula C=4R, where is the permittivity of open space, may be used to determine the capacitance of a merged mercury drop, assuming it is an isolated sphere. The capacitance is around 1.68 pF with R = 5.61 mm.
The formula C=4R, where R is the drop's radius and is the permittivity of free space, may be used to determine the capacitance of a merged mercury drop. As the capacitance of an isolated sphere is exactly proportional to its radius, the capacitance produced by the merger of two drops with similar radii is equal to the total of the capacitances of the individual drops. Given that the radius of the combined drop in this instance is R = 5.61 mm, the capacitance can be estimated using the formula C = 4(8.85 x 10-12 F/m) (5.61 x 10-3 m)2, yielding a capacitance of around 1.68 pF.
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A small grinding wheel has a moment of inertia of 4. 0×10−5 kg⋅m2
k
g
⋅
m
2. What net torque must be applied to the wheel for its angular acceleration to be 150 rad/s2
r
a
d
/
s
2
?
A net torque of [tex]6.0×10^−3 N⋅m[/tex] is sufficient to produce the desired angular acceleration of [tex]150 rad/s^2[/tex].
The net torque required to produce an angular acceleration in a rotating object can be calculated using the formula: net torque = moment of inertia × angular acceleration In this case, the moment of inertia of the grinding wheel is given as 4.0×10^−5 kg⋅m^2 and the angular acceleration required is 150 rad/s^2.
Therefore, the net torque required can be calculated as: net torque = [tex](4.0×10^−5 kg⋅m^2) × (150 rad/s^2) = 6.0×10^−3 N⋅m[/tex]To explain this result, we need to understand the relationship between torque and angular acceleration. Torque is the rotational equivalent of force and it is defined as the product of force and the perpendicular distance between the line of action of the force and the axis of rotation.
When a torque is applied to a rotating object, it produces an angular acceleration in the object, which is a measure of how quickly the object's rotational speed changes.
The moment of inertia of an object is a measure of its resistance to changes in its rotational motion. It depends on the object's mass distribution and the distance of each element of mass from the axis of rotation. Objects with larger moments of inertia require more torque to produce a given angular acceleration than objects with smaller moments of inertia.
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In outer space will a liquid in a beaker exert a pressure on the bottom or on the sides of a beaker?
Answer:
Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.
a mass of 0.3 kg hangs motionless from a vertical spring whose length is 1.08 m and whose unstretched length is 0.42 m. next the mass is pulled down to where the spring has a length of 1.21 m and given an initial speed upwards of 1.2 m/s. what is the maximum length of the spring during the motion that follows?
The maximum length of the spring (Lmax) is about 1.25 meters.
What is the maximum length of spring?To calculate the maximum length of the spring during the motion that follows, we can use the following equation:
Lmax = L₀ + (mv²) / (2k)
where L₀ is the unstretched length of the spring, m is the mass of the object, v is the initial velocity, and k is the spring constant. In this case, L₀ = 0.42 m, m = 0.3 kg, v = 1.2 m/s, and k = 39 N/m.
The maximum length of the spring is:
Lmax = 0.42 + (0.3 × (1.2)²) / (2 × 39) = 1.25 meters
Therefore, the maximum length of the spring during motion is 1.25 meters.
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three forces applied to a trunk that moves leftward by 3.010 m over a frictionless floor. The force magnitudes are F1 = 5.86 N, F2 = 9.180 N, and F3 = 3.850 N, and the indicated angle is θ = 67.8°. During the displacement, what is the net work done on the trunk by the three forces? (Note that there are other forces acting on the block, but we only care about the net work done by these three forces.) And by how much does the kinetic energy of the trunk increase (enter a positive value) or decrease (negative value)?
The kinetic energy of the trunk increases by ½ mvf² = ½ m(10.65 m/s)²= 71.44 J during the displacement.
Net work = ΔK
W = Fd cosθ
W1 = F1d cosθ = (5.86 N)(3.010 m) cos(67.8°) = 6.99 J
W2 = F2d cosθ = (9.180 N)(3.010 m) cos(67.8°) = 10.97 J
W3 = F3d cosθ = (3.850 N)(3.010 m) cos(67.8°) = 4.58 J
Net work = W1 + W2 + W3 = 6.99 J + 10.97 J + 4.58 J = 22.54 J
Therefore, the net work done on the trunk by the three forces is 22.54 J.
ΔK = ½ mvf² - ½ mvi²
Since the trunk moves a distance of 3.010 m and is initially at rest, we can use the equation:
vf² = 2ad
where a is the acceleration of the trunk, which is given by:
a = ΣF / m
where ΣF is the net force on the trunk, which we can find using:
ΣF = F1 + F2 + F3
ΣF = (5.86 N + 9.180 N + 3.850 N) = 18.89 N
Therefore, the acceleration of the trunk is:
a = ΣF / m = 18.89 N / m
Since the trunk moves leftward, the acceleration is also leftward, so we can use a negative value for a.
Substituting the values for a and d, we get:
vf² = -2(-18.89 N / m)(3.010 m) = 113.51 (m/s)²
Taking the square root, we get:
vf = 10.65 m/s
Therefore, the change in kinetic energy of the trunk is:
ΔK = ½ mvf² - ½ mvi² = ½ m(10.65 m/s)²- 0 = ½ mvf²
Kinetic energy is a type of energy that an object possesses by virtue of its motion. It is defined as the energy an object has due to its motion and is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the object and v is its velocity.
Kinetic energy is a scalar quantity and has units of joules in the International System of Units (SI). It is a fundamental concept in physics and is used to describe many physical phenomena, including the motion of particles, the behavior of gases, and the motion of waves. In many cases, kinetic energy can be transformed into other forms of energy. For example, when a ball is thrown upwards, its kinetic energy is gradually converted into gravitational potential energy as it moves higher and higher.
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on a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . part a how much current flows through the panel? express your answer with the appropriate units.
On a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . the current flows through the panel is 3.235 A.
The amount of current flowing through the panel can be calculated using Ohm's Law (I = p/v). When working with the formula, Power = Voltage × Current (P = V × I), one can determine the current that flows through the panel by rearranging the formula to:
Current = Power/Voltage (I = P/V).
The calculation of the current (I) is given as follows: I = P/V = 55 W / 17 V = 3.235 A.
Therefore, 3.235 A of current flows through the solar panel on a sunny day.
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At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula c=α+βT+δT−2c=α+βT+δT−2, where α=349J/kg⋅K,β=0.107J/kg⋅K2α=349J/kg⋅K,β=0.107J/kg⋅K2, and δ=4.58×105J⋅kg⋅Kδ=4.58×105J⋅kg⋅K. How much heat is needed to raise the temperature of a 2.00-kg piece of copper from 20∘C to 250∘C20∘C to 250∘C?
The amount of heat required to raise a 2.0 kg piece of copper from [tex]20^\circ C[/tex] to [tex]250^\circ C[/tex] using the formula [tex]c = \alpha + \beta T + \delta T^{-2}[/tex] is [tex]1.96 \times 10^{8} J[/tex].
The formula for computing the amount of heat that is required to raise the temperature of an object is expressed as:
Q = mc∆T
Where:
Q: the amount of heat needed in joules
m: the mass of the substance in kilograms
c: the specific heat capacity of the substance in joules per kilogram per kelvin
∆T: the change in temperature in kelvin
We can use the equation to calculate the amount of heat needed to raise the temperature of a 2.00-kg piece of copper from 20∘C to 250∘C. However, it's important to note that we need to first convert the given temperatures from Celsius to Kelvin.
[tex]20^\circ C + 273 = 293 K[/tex] (Initial temperature)
[tex]250^\circ C + 273 = 523 \ K[/tex] (Final temperature)
We can now substitute the values into the formula, Q = mc∆T to get the amount of heat required.
[tex]Q = mc\Delta T[/tex]
[tex]Q = (2.00 \ kg) (\alpha + \beta T + \delta T^{-2}) (\Delta T)[/tex]
[tex]Q = 2.00 kg (\alpha\Delta T + \beta \Delta T^2 + \delta \Delta T^3)[/tex]
[tex]Q = 2.00 kg [(\alpha(523 \ K - 293 \ K) + \beta (523 K^2 - 293\ K^2) + \delta(523 \ K^3 - 293 \ K^3)][/tex]
[tex]Q=2.00 kg [(349 J/kgK \times 230\ K) + (0.107 J/kgK^2 \times 230 K^2) + (4.58 \times 10^5 JkgK \times 230 K^3)][/tex]
[tex]Q = 2.00\ kg [80270\ J + 24.61\ J + 9.8 \times 10^{7} \ J][/tex]
[tex]Q = 1.96 \times 10^{8} J[/tex]
Therefore, the amount of heat that is required to raise a 2.00-kg piece of copper from [tex]20^\circ C[/tex] to 250° C is [tex]1.96 \times 10^{8} J[/tex].
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A charge, q1 = +4. 00 MC, is at the origin, and a second charge, 92 =
-6. 00 MC, is on the x-axis 0. 300 m from the origin. Find the electric field at a point "+P" on the y-axis 0. 800 m from the origin. What is the net force on "p" (magnitude and direction)
The electric field at a point "+P" on the y-axis 0. 800 m from the origin is 53.3 N/C. The net force on "p" (magnitude and direction) is 5.33 x 10^-5 N.
To find the electric field at point "p" on the y-axis, we can use Coulomb's law and the principle of superposition.
First, let's find the electric field contribution at point "p" due to the charge q1 at the origin. We can use Coulomb's law for point charges to find the electric field contribution:
E = k * q / r²
where k is Coulomb's constant, q is the charge, and r is the distance from q to point "p". In this case, r is simply the distance from the origin to point "p", which is 0.8 m. Plugging in the values:
E1 = (9.0 x 10⁹ N*m²/C²) * (+4.00 x 10-⁶ C) / (0.8 m)²
E1 = 18.0 N/C (upwards on the y-axis)
Similarly, the electric field contribution at point "p" due to the charge q2 on the x-axis and at a distance r2 can be calculated Using the Pythagorean theorem, we can find this distance:
r2 = √[(0.3 m)² + (0.8 m)²] = 0.854 m
Plugging in the values:
E2 = (9.0 x 10⁹ N*m²/C²) * (-6.00 x 10-⁶ C) / (0.854 m)²
E2 = 50.6 N/C (at an angle of arctan(0.8/0.3) = 69.4 degrees below the negative x-axis)
To find the total electric field at point "p", we add the contributions from q1 and q2 using vector addition:
Etotal = E1 + E2
Using the component method, we can find the magnitude and direction of the total electric field:
|Etotal| = √[(E_total,x)² + (E_total,y)²]
= √[(-18.0 N/C)² + (50.6 N/C)²]
= 53.3 N/C
θ = arctan[(E_total,y) / (E_total,x)]
= arctan[(50.6 N/C) / (-18.0 N/C)]
= -69.2 degrees
Therefore, the magnitude of the net force on a +1.00 C test charge placed at point "p" is,
Fnet = qtest * |E_total| = (+1.00 x 10^-6 C) * (53.3 N/C) = 5.33 x 10^-5 N
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consider two planets in space that gravitationally attract each other. if the masses of both planets are doubled, and the distance between them is also doubled, then the force between them is group of answer choices four times as much. half as much. twice as much. remains the same. one quarter.
If the masses of two planets in space that gravitationally attract each other are doubled and the distance between them is also doubled, then the force between them remains the same.
To determine the force between two planets, we use the formula:
F = Gm1m2/r^2
where F is the force of gravitational attraction between the two planets, G is the gravitational constant, m1 and m2 are the masses of the planets, and r is the distance between them.
If both masses are doubled and the distance between them is also doubled, the new force between them can be calculated as follows:
F' = G(2m1)(2m2)/(2r)^2
Simplifying this expression, we get:
F' = Gm1m2/r^2
which is the same as the original force between the planets.
Therefore, if the masses of two planets are doubled and the distance between them is also doubled, the force between them remains the same.
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a researcher is studying the distribution of auxin in roots and stems exposed to sunlight. he notices that more auxin collects in the sides of stems and roots that are not exposed to light. why?
The researcher's observation that more auxin collects in the sides of stems and roots that are not exposed to light is likely due to the phenomenon of phototropism.
In the process of phototropism, light influences the direction and rate of growth of plant cells. In particular, light induces the cells on one side of a stem or root to create less auxin than the cells on the shaded side. Less auxin is produced on the lighted side and more auxin is produced on the shaded side as a result. The hormone auxin is essential for controlling the growth and development of plants. Auxin generally promotes cell growth and elongation at greater concentrations while inhibiting cell elongation at lower concentrations. Since the cells on the lighted side of the stem or root will contain less auxin when there is light.
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Imagine sitting on a merry-go-round and riding along as it spins. Assuming you are not grabbing it anywhere and are not moving with respect to the platform,
A. static friction (directed inwards) causes you to accelerate.
B. you are not accelerating because you aren't moving on the platform.
C. static friction (directed outwards) causes you to accelerate.
D. sliding friction makes you accelerate inwards.
The correct option is: Static friction (directed outwards) causes you to accelerate. (Option C)
When you sit on a merry-go-round, you are not moving relative to the platform. Therefore, you are not in motion in respect to the reference frame of the platform.
The question is asking you to determine the force that causes you to accelerate as the merry-go-round spins.
Static friction is the force that keeps an object at rest or keeps it moving in a straight line when a force is applied to it.
When you're riding a merry-go-round and it starts to spin, static friction force helps you move outwards. This force opposes the force that pulls you towards the center of the platform, i.e., centripetal force.
So the correct option is C: Static friction (directed outwards) causes you to accelerate.
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A small mass rests on a horizontal platform which vibrates vertically in simple harmonicmotion with period 0.50 s.(a) Find the maximum amplitude of the motion which will allow the mass to stay in contactwith the platform throughout the motion.The maximum acceleration that will allow the object to remain in contact with theplatform at all times is when amax = g = 9:81 m/s.But amax = !222A = (2¼=T )A ) 9:81 = (2¼=0:5)2A = 158A ) A = 0:062 m
The maximum amplitude of the motion which will allow the mass to stay in contact with the platform is 0.062 m.
This can be calculated by using the equation amax = (2π/T)2A, where A is the maximum amplitude, and T is the period of the motion. In this case, T is 0.50 s, and g (the acceleration due to gravity) is 9.81 m/s2, so we can calculate A:
A = (2π/T)2g = (2π/0.50)2 × 9.81 = 158 × 9.81 = 1543.38
Therefore, A = 1543.38/158 = 9.81 m/s2 = 0.062 m.
Alternatively: given,T = 0.50 s,The acceleration due to gravity, g = 9.81 m/s²Maximum acceleration, amax = g = 9.81 m/s². The maximum acceleration that will allow the object to remain in contact with the platform at all times is when amax = !222A = (2π/T )A ) 9.81 ...(1)From the equation (1), we get 158 A = 9.81 (2π/0.50)A = (9.81 (2π/0.50))/158 = 0.062 m. Therefore, the maximum amplitude of the motion which will allow the mass to stay in contact with the platform throughout the motion is 0.062 m.
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Two 4.4 kg bodies, A and B, collide. The velocities before the collision are A = (28i + 27j) m/s and B = (9.8i + 1.8j) m/s. After the collision, 'A = (3.7i + 3.2j) m/s. What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?
Answer:jfnvufhdfiprhfpiurgh8rhvjm vjfnb
Explanation:
in the bohr model of the hydrogen atom, an electron in the lowest energy state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 1011 m. what is the effective current associated with this orbiting electron?
The effective current associated with an electron in the lowest energy state of the hydrogen atom in the Bohr model is 6.63×10-7 A.
To calculate the effective current, we can use the formula:
I = qv/T
where I is the effective current, q is the charge of the electron, v is its velocity, and T is the time period of its circular orbit.
In the lowest energy state of the hydrogen atom, the electron is in a circular orbit with a radius of 5.29×1011 m and a speed of 2.19×106 m/s. The time period of the orbit can be calculated using the formula for centripetal acceleration:
a = v^2/r
F = ma = (mv^2)/r
F = kQq/r^2
mv^2/r = kQq/r^2
T = 2pir/v
where F is the electrostatic force between the electron and the proton in the nucleus, k is Coulomb's constant, Q is the charge of the nucleus, q is the charge of the electron, m is the mass of the electron, and r is the radius of the orbit.
Substituting the given values, we get:
T = 2pi(5.29×10^(-11) m)/(2.19×10^6 m/s) = 2.42×10^(-16) s
Using the charge of an electron, q = -1.6×10^(-19) C, and the velocity calculated above, we get:
I = qv/T = (-1.6×10^(-19) C)(2.19×10^6 m/s)/(2.42×10^(-16) s) = -6.63×10^(-7) A
The negative sign indicates that the effective current is in the opposite direction of the electron's motion.
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the bottom of the tank is 30 feet in diameter and groundwater level is 10 ft above the bottom of the tank. how much force is being exerceted on the bottom of the tank?
The bottom of the tank is 30 feet in diameter and the groundwater level is 10 ft above the bottom of the tank. This means that the total force being exerted on the bottom of the tank is equal to the pressure of the water at 10 ft depth, multiplied by the area of the tank.
The pressure of water at 10 ft depth can be calculated using the formula: Pressure = Density of Water x Gravity x Depth
Therefore, the force exerted on the bottom of the tank is:
Force = Density of Water x Gravity x Depth x Area of the TankWhere:
Density of Water is 62.4 lb/ft³Gravity is 32.2 ft/s2, Depth is 10 ftArea of the Tank is 706.5 ft²Therefore, the total force exerted on the bottom of the tank is: Force = 62.4 x 32.2 x 10 x 706.5 = 1358561.2 lbf.
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an arrow leaves a bow with a speed of 42 m/s. its velocity is reduced to 34 m/s by the time it hits its target. how much distance did the arrow travel over if it were in the air for 2.4 seconds?
The distance did the arrow travel over if it were in the air for 2.4 seconds is 100.8 meters.
What is the distance?An arrow leaves a bow with a speed of 42 m/s. Its velocity is reduced to 34 m/s by the time it hits its target. And the arrow traveled in the air for 2.4 seconds.
To find the distance traveled by the arrow, we can use the following formula:
S = v₀t + 1/2at²
where, S = distance traveled v₀ = initial velocity = 42 m/s, t = time taken = 2.4 s, a = acceleration = ? u = final velocity = 34 m/s.
As per the question, the arrow is traveling through the air, so the acceleration is due to gravity, which is equal to 9.8 m/s².So, a = 9.8 m/s². Now, we can substitute the given values in the above formula:
S = 42 m/s × 2.4 s + 1/2 × 9.8 m/s² × (2.4 s)²
S = 100.8 m.
The arrow traveled approximately 100.8 meters in the air.
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A car travelling at 22.4 m/s skids to a stop in 2.55s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
Approximately [tex]28.6\; {\rm m}[/tex].
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle, and let [tex]v[/tex] denote the velocity of the vehicle after skidding. It is given that the initial velocity was [tex]u = 22.4\; {\rm m\cdot s^{-1}}[/tex]. Since the vehicle skidded to a stop, [tex]v = 0\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]t[/tex] denote the duration of the skid. It is given that [tex]t = 2.55\; {\rm s}[/tex].
Under the assumption that acceleration is constant, SUVAT equations will apply.
Specifically, the SUVAT equation [tex]x &= (1/2)\, (u + v)\, t[/tex] will be satisfied. In this equation, the displacement of the vehicle is equal to average velocity times duration. This equation allows the displacement [tex]x[/tex] to be found from [tex]u[/tex], [tex]v[/tex], and [tex]t[/tex] without knowing the exact value of acceleration:
[tex]\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \left(\frac{22.4 + 0}{2}\; {\rm m\cdot s^{-1}}\right)\; (2.55\; {\rm s}) \\ &\approx 28.6\; {\rm m}\end{aligned}[/tex].
explain why hydrogen is not a dominant component of the atmospheres of the terrestrial planets, even though it is the most common element in the universe and in the solar system.
Hydrogen is not a dominant component of the atmospheres of the terrestrial planets, even though it is the most common element in the universe and solar system because it is a volatile and low molecular weight element.
What is Hydrogen?Hydrogen element is not a dominant component of the atmospheres of the terrestrial planets, even though it is one of the most common element in the universe and in the solar system as well because the planets were formed through the accretion.
Accretion is the process by which the small particles combine into progressively larger bodies which become planets. As a result of this, the process left behind the lighter gas elements such as hydrogen and helium. The gas molecules of hydrogen atom are lighter than the rest of the chemical elements, therefore the gravity of the terrestrial planets is not sufficient to hold them. As a result, hydrogen escapes from the planet's atmosphere, leaving behind the heavier elements like carbon dioxide, nitrogen, and oxygen in their atmospheres.
Therefore, although hydrogen is the most common element in the universe and in the solar system, it is not a major constituent of the terrestrial planets.
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How does the star formation in spirals compare to the star formation of elliptical galaxies?
The spiral galaxies are characterized by the arms winding out from a central nucleus while the elliptical galaxies are characterized by their lack of structure or a central bulge.
Star formation refers to the process by which dense areas within molecular clouds in interstellar space, typically lasting tens of millions of years, form newborn stars. It takes a long time for stars to form, and this process is not well understood.
In comparison to spiral galaxies, elliptical galaxies have low star formation.
Furthermore, elliptical galaxies are made up of stars with a wide range of ages, indicating that the star formation process was rapid and early on in their history.
Spiral galaxies have more gas and dust in their disks than elliptical galaxies, and these are the sites of intense star formation.
The arms are believed to be regions of higher density of stars and interstellar material, as well as more significant gravitational interactions among stars, gas, and dust than in the rest of the disk.
Thus, spiral galaxies are sites of ongoing star formation while elliptical galaxies are mainly populated by old and evolved stars that no longer form. Therefore, spiral galaxies have a higher rate of star formation than elliptical galaxies.
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you live on an island in the pacific. an earthquake of magnitude 8.5 off the coast of japan, 8000 km away, generates a tsunami with a wavelength of 200 km. the average water depth between your island and japan is 4900 m. if a tsunami warning is issued for your island, how many hours will you have before the waves arrive?
If a tsunami warning is issued for the island, they will have approximately 11.7 hours before the waves arrive.
What is Magnitude?
Magnitude is a measure of the strength or intensity of a physical quantity or phenomenon, such as an earthquake or a sound wave. It is often expressed using a numerical scale, with higher values indicating greater strength or intensity. In the case of earthquakes, magnitude is typically measured using the Richter scale or the moment magnitude scale, which take into account the amplitude of seismic waves and the energy released by the earthquake.
To calculate the time it takes for a tsunami to travel from Japan to the island, we can use the following formula:
t = (2 * pi * d) / g * ln(1 + sqrt(h/d))
where t is the time it takes for the tsunami to travel, d is the average water depth, h is the wave height, and g is the acceleration due to gravity (9.8 m/s^2).
Magnitude of the earthquake: 8.5
Wavelength of the tsunami: 200 km = 200,000 m
Average water depth: 4,900 m
To calculate the wave height, we can use the following formula:
h = (M / 5) * (D / 10)^1/2
where M is the magnitude of the earthquake and D is the distance between the earthquake epicenter and the observation point (in this case, the island). Note that this formula is an approximation and may not be accurate for all cases.
Using the given values, we get:
D = 8,000 km = 8,000,000 m
h = (8.5 / 5) * ((8,000,000 / 10)^1/2) = 2,738.6 m
Substituting these values into the formula for t, we get:
t = (2 * pi * 4,900) / 9.8 * ln(1 + sqrt(2,738.6/4,900)) = 11.7 hours
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in addition to hundreds of smaller objects they have been discovering in the kuiper belt recently, astronomers were surprised to find
In addition to hundreds of smaller objects they have been discovering in the Kuiper Belt recently, astronomers were surprised to find dwarf planet Eris.
The first object that was bigger than Pluto was Eris. The initial estimate of Eris' size was 1,240 miles (2,000 kilometers) in diameter. It was later discovered to be a bit smaller, with a diameter of 1,163 miles (1,864 kilometers). Its moon, Dysnomia, was also discovered.Eris' orbit is far more eccentric than Pluto's, ranging from 38 to 97 astronomical units (AU) from the Sun.
Eris takes 557 Earth years to orbit the Sun. Despite the fact that Pluto's path also varies in shape, it is always closer to the Sun than Eris. Pluto and Eris were both discovered in the early 21st century, in 1930 and 2005, respectively. Because it was the largest known body in the Kuiper Belt, Pluto was formerly classified as the Solar System's ninth planet. Following the discovery of Eris and other trans-Neptunian objects, Pluto was reclassified as a dwarf planet.
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When a ball bounces against a wall there will be large change in velocity in short period of time. This means the ____ is large, hence the net ___ must be proportionately large as well.
A change in velocity in short period of time means the acceleration is large, hence the net force must be proportionately large as well.
What is a force?A force is a physical quantity that induces a body to undergo an alteration in speed, direction of motion, or shape. A force can be classified as a push or a pull. When forces are equal, the forces are balanced and the object is not moving. Otherwise, if the forces are not equal, making it unbalanced will not give the object any movement.
The force that induces the change in the speed or direction of an object is referred to as a net force. The net force is equal to the product of the mass of the object and its acceleration. Newton (N) is the unit of measurement for force.
When a ball bounces against a wall, there will be a large change in velocity in a short period of time. This means the acceleration is large, hence the net force must be proportionately large as well.
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Galena's specific gravity is 7.5, that of quartz 2.65, and that of liquid mercury 13.6. Given equal-sized samples (volumes) of galena and quartz, which will feel heavier? Choose one: A. galena B. The same volume of water will feel heavier than both of them. C. They will feel about equal. D. quartz
Given equal-sized samples (volumes) of galena and quartz, the Galena sample will feel heavier because of its higher specific gravity. Thus, the correct option is A.
What is the Specific gravity of a substance?Specific gravity is the ratio of the density of a substance to the density of a standard substance in physics. It's typically applied to liquids and solids, but it may also be applied to gases. The most often utilized standard material for liquids and solids is water at 4°C. A substance's specific gravity is dimensionless and is often represented by the Greek symbol ρ.
Relative Density of the given substances:
Galena's specific gravity is 7.5, Quartz's specific gravity is 2.65, and Liquid mercury's specific gravity is 13.6. An object with a specific gravity greater than 1 sinks in water, while one with a specific gravity less than 1 floats in water. The specific gravity of water is 1.0. An object with a specific gravity greater than 1 sinks in water, while one with a specific gravity less than 1 floats in water.
We can conclude from the values above that liquid mercury is heavier than galena, which is in turn heavier than quartz. Therefore, since both quartz and galena are being measured with equal sizes or volumes, galena will feel heavier than quartz.
Therefore, the correct option is A.
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spherical capacitor contains a charge of 3.20nCwhen connected to a potential difference of250V. If its plates are separated by vacuum and theinner radius of the outer shell is 4.60cm.
A) Calculate the capacitance.
B) Calculate the radius of the inner sphere.
C) Calculate the electric field just outside the surface of theinner sphere.
A) The capacitance of the spherical capacitor is 1.45 pF (picofarads), B) The radius of the inner sphere is 3.60 cm. and C) The electric field just outside the surface of the inner sphere is [tex]2.36 * 10^6 V/m[/tex] (volts per meter).
To calculate the capacitance, we can use the formula C = Q/V, where Q is the charge and V is the potential difference. Plugging in the values, we get [tex]C = (3.20 * 10^{-9} C)/(250 V) = 1.28 * 10^{-11} F[/tex].
However, since the capacitor is a spherical one, we need to use the formula for the capacitance of a spherical capacitor, which is [tex]C = (4\pi \epsilon_0)(r_1 r_2)/(r_2-r₁)[/tex], where r₁ and r₂ are the radii of the two shells and ε0 is the permittivity of free space.
Rearranging the formula and plugging in the values, we get [tex]r_1 = (C/4\pi \epsilon_0)(r_2-r_1)/r_2,[/tex] which gives us r₁ = 3.60 cm.
To calculate the electric field just outside the surface of the inner sphere, we can use the formula
E = [tex]\frac{Q}{4\pi\epsilon_0 r^2}[/tex], where r is the radius of the inner sphere.
Plugging in the values, we get [tex]E = (3.20 * 10^{-9} C)/(4\pi\epsilon_0(0.0460 m)^2) = 2.36 * 10^6 V/m.[/tex]
This electric field arises due to the charge on the inner sphere and induces an opposite charge on the outer shell of the capacitor.
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a particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. the particle's position at t0=0s is x0 = -5.40 m . at t1 = 2.00 s , the particle is at x1 = 5.80 m .
A particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. The particle's position at t0=0s is x0 = -5.40 m. At t1 = 2.00 s , the particle is at x1 = 5.80 m. The value of k is 2.80 m/s2.
The given equation describes the velocity of a particle in terms of a constant, k, and time, t. The velocity, vx, is given in m/s. The initial position of the particle at t0=0s is x0=-5.40 m, and at t1=2.00 s the particle is at x1=5.80 m. To find the value of the constant k, we can solve the equation for the change in velocity Δvx.
Δvx = vx1 – vx0 = k(t12 – t02)
Δvx = 5.80 – (-5.40) = 11.20 m/s
k = (11.20 m/s) / (2.002 s2) = 2.80 m/s2
Now that we have found the value of the constant k, we can use it to find the velocity of the particle at any time t. For example, at t2=4.00 s the velocity of the particle is vx2=11.20 m/s. This can be calculated using the equation vx2 = k(t22) = 2.80(4.002) = 11.20 m/s.
From the velocity equation, we can also calculate the position of the particle at any time t. The position of the particle at t2=4.00 s is x2= 11.20(4.00) = 44.80 m. We can also calculate the position of the particle at any other time t, by simply substituting in the corresponding value of t into the equation.
In conclusion, the equation vx = kt2 describes the velocity of a particle in terms of a constant, k, and time, t. Using this equation, we can calculate the velocity and position of the particle at any given time.
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Complete Question:
A particle’s velocity is described by the function vx = [tex]kt^2m/s[/tex], where k is a constant and t is in s. The particle’s position at [tex]t_0[/tex] = 0s is [tex]x_0[/tex] = -5.40 m. At [tex]t_1[/tex] = 2.00 s, the particle is at [tex]x_1[/tex] = 5.80 m. Determine the value of the constant k. Be sure to include the proper units
Taking the following list on an item-by-item basis (i.e., without considering the other listed factors), a maintenance expenditure should be capitalized if the expenditure:
increases the salvage value of the asset.
extends the useful life of the asset.
A maintenance expenditure should be capitalized if it increases the salvage value of the asset or extends the useful life of the asset.
An expenditure is a payment made in return for a product or service. Capital expenditure is money spent by a company on long-term assets like equipment and buildings.
Capitalizing refers to recording a cost or expense on the balance sheet for a future period rather than recognizing it immediately in the current period.
Capitalizing expenditure means the company will recognize the expenditure as an asset, which will be amortized over its useful life as opposed to expenses in the current period.
Therefore, a maintenance expenditure should be capitalized if the expenditure increases the extends the useful life of the asset.
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You are the process engineer at Corvallis Automobiles Inc., and you have received an order to turn a cylindrical bar on an engine lathe to the dimensions specified in Fig. 1. For this order you will use cylindrical bar stock that is 48-inches long and 4-inches in diameter. The 48-inch length bar will be chucked in the lathe and supported at the opposite end using a live center. You are planning to complete the operation in one pass using a cutting speed of 400 ft./min. and a feed of 0.010 in./rev. Determine the following: a) The required depth of cut (in inches) b) The material removal rate (in cubic inches per minute)
c) The time required to complete the cutting pass (in minutes)
a. the depth of cut is 0.625 inches.
b. the material removal rate is 0.003125 cubic inches per minute.
c. the time required to complete the cutting pass is 20 minutes.
How do we calculate?a) The required depth of cut can be determined by :
DOC = (4 in - 2.75 in)/2 = 0.625 in
Therefore, the depth of cut is 0.625 inches.
b) The material removal rate can be found by applying:
MRR = DOC x Width of cut x Feed rate
assuming we are using a standard carbide insert tool with a width of cut of 0.5 inches.
MRR = 0.625 in x 0.5 in x 0.010 in/rev = 0.003125 cubic inches per minute
c) The time required to complete the cutting pass is determined by:
Time = Length of cut / (Cutting speed x Width of cut x Feed rate)
Time = 48 in / (400 ft/min x (0.5 in) x (0.010 in/rev) x (1/12 ft/in)) = 20 minutes
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A car has an intial velocity of 50 km hr after 5 h, its final velocity is 70 km hr. solve for the car acceleration
Answer:
4 km/hr^2
Explanation:
We can use the formula for acceleration:
a = (v_f - v_i) / t
where:
a = acceleration
v_f = final velocity
v_i = initial velocity
t = time taken
Substituting the given values, we get:
a = (70 km/hr - 50 km/hr) / 5 hr
a = 20 km/hr / 5 hr
a = 4 km/hr^2
