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3 years ago

0.5kg ball moving at a speed of 3m/s rolls up a hill what is thet height

Physics

1 answer:

Annette [7]3 years ago

5 0

Answer:

0.459 m.

Explanation:

To calculate the height, use 3rd equation of motion as

$v^2-u^2=2gh$

Here v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and h is the height.

As ball moving up so acceleration due to gravity becomes negative i.e -g and at maximum height v = 0.

Therefore equation becomes,

$u=\sqrt{2gh}$

Given u = 3 m/s and g = 9.8 m/s^2.

Substitute the given values, we get

$(3m/s)^2=2\times9.8m/s^2h$

h = 0.459 m

Thus, the height of hill is 0.459 m.

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3 years ago

Read 2 more answers

A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls

anastassius [24]

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

$KE_r = \dfrac{1}{2}I\omega^2$

Moment of inertia of coin

$I = \dfrac{1}{2}MR^2$

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$KE_t = \dfrac{1}{2}MR^2\omega^2$

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KE_r + KE_t = PE

$\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh$

$\dfrac{3}{4}R^2\omega^2=gh$

$\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h$

h = 0.0259 m

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4 years ago

In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 sec

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Answer:

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Time, t = 0.4 s

We need to find the magnitude of the average horizontal force (lb) exerted on the car during the impact. It can be calculated as :

$F=m\times \dfrac{v-u}{t}\\\\F=1270.059\times \dfrac{-0.67056 -2.2352 }{0.4}\\\\F=9226.21\ N$

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