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3 years ago

Momentum of 0.5 kg moving ball with the velocity of 5 m / s

Physics

2 answers:

insens350 [35]3 years ago

6 0

Answer:

$\boxed{ \bold{ \huge{\boxed{ \sf{2.5 \: kg \: m \: / \: s}}}}}$

Explanation:

Mass ( m ) = 0.5 kg

Velocity ( v ) = 5 m / s

Momentum ( p ) = ?

Finding the momentum

$\boxed{ \sf{momentum \: ( \: p \: ) \: = \: mass \: \times \: velocity}}$

⇒ $\sf{momentum \: = \: 0.5 kg \times 5 \: meter \: \: per \: second}$

⇒ $\sf{momentum = 2.5 \: kg \: m \: / \: s}$

Hope I helped!

Best regards! :D

ankoles [38]3 years ago

3 0

Answer:

2.5

Explanation:

Momentum = Mass × Velocity

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⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)

GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

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(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

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andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

When a similar person of the same mass is standing at a distance of 4 meters:

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The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

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Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

Explanation:

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Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal

Dmitry_Shevchenko [17]

Answer:

$a = \dfrac{4\pi^2R}{T^2}$

Explanation:

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$a = \omega^2 R$

where $\omega$ is the angular velocity and $R$ is the radius.

Angular velocity is related to the period, T, by

$\omega=\dfrac{2\pi}{T}$

Substitute into the previous formula.

$a = (\dfrac{2\pi}{T})^2 R$

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3 years ago

Momentum of 0.5 kg moving ball with the velocity of 5 m / s ​

Momentum of 0.5 kg moving ball with the velocity of 5 m / s