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2 years ago

Is kicking a ball a conservation of momentum ?

Physics

1 answer:

Katarina [22]2 years ago

7 0

Answer:

Yes, When the momentum and kinetic energy are conserved, as they are approximately conserved when kicking a ball, there is no loss in momentum and energy before or after the collision.

Hope this helps!

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A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

2 years ago

Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

4 0

2 years ago

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

$I_i$ = Initial angular momentum = $275\ kgm^2$

$\omega_i$ = Initial angular velocity = 14 rev/min

$I_f$ = Final angular momentum

$\omega_f$ = Final angular velocity

Here the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s$