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3 years ago

7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec

Physics

1 answer:

lys-0071 [83]3 years ago

8 0

The time elapsed is 9 seconds

Explanation:

The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:

$v=u+at$

where :

v is the final velocity of the ball

u is the initial velocity

a is the acceleration

t is the time elapsed

For the ball in this problem, we have:

u = 3 m/s is the initial velocity

v = 34.5 m/s is the final velocity

$a=3.5m/s^2$ is the acceleration

Solving for t, we find the time taken for this change in velocity:

$t=\frac{v-u}{a}=\frac{34.5-3}{3.5}=9 s$

Learn more about acceleration:

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4 years ago

A projectile is fired from ground level with an initial speed of 650 m/sec and an angle of elevation of 30 degrees. Use that the

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Answer:

(a) The range of the projectile is 37,336.3 m

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Explanation:

Given;

initial velocity of the projectile, u = 650 m/s

angle of projection, θ = 30⁰

(a) The range of the projectile is calculated as;

$R = \frac{u^2sin 2\theta}{g} \\\\R = \frac{(650^2)Sin (2\times 30^0)}{9.8} \\\\R = 37,336.3 \ m$

(b) The maximum height of the projectile is calculated as;

$H = \frac{u^2sin^2\theta}{2g} \\\\H = \frac{(650^2)(Sin \ 30^0)^2}{2\times 9.8} \\\\H = \frac{(650^2)(0.5)^2}{19.6} \\\\H = 5,389.03 \ m$

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3 years ago

What is the volume of a cube with length = 3 centimeters, width = 3

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Answer:

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Explanation:

l*w*h

3*3*3= 27

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3 years ago

A 0.140 kg stone rests on a frictionless, horizontal surface. A bullet of mass 8.50 g , traveling horizontally at 320 m/s , stri

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Answer:

a) $v = 34.607\,\frac{m}{s}$ (Positive), b) $e = 0.803$ . The collision is not perfectly elastic.

Explanation:

a) The collision can be described by the Principle of Momentum Conservation and Principle of Energy Conservation:

$(0.140\,kg)\cdot (0\,\frac{m}{s} ) + (0.0085\,kg)\cdot (320\,\frac{m}{s} ) = (0.0085\,kg)\cdot (-250\,\frac{m}{s} ) + (0.140\,kg)\cdot v$

The final velocity of the rock is:

$v = 34.607\,\frac{m}{s}$

b) The coefficient of restitution is the best criterion to distinguish elastic collsions from inelastic collisions, such criterion is the ratio of final energy of the system to initial energy of the system:

$e = \frac{\frac{1}{2}\cdot [(0.140\,kg)\cdot (34.607\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (-250\,\frac{m}{s} )^{2}] }{\frac{1}{2}\cdot [(0.140\,kg)\cdot (0\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (320\,\frac{m}{s} )^{2}] }$