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3 years ago

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A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What

1 answer:

enot [183]3 years ago

4 0

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

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A table tennis ball with a mass of 0.003 kg and a soccer ball with a mass of 0.43 kg or both Serta name motion at 16 M/S calcula

poizon [28]

Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= ( kg-m/s ) I use the subscript "t" for tennis. Momentum of Soccer ball Ps= 0.43 x 13m/s = ( km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.

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3 years ago

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What is kinetic energy?

Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

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3 years ago

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A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

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3 years ago

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A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

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Amanda [17]

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