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1 year ago

A ball rolls up a hill, going from a speed of 8.09 m/s to a speed of 5.87 m/s in 9.17 seconds. What is its acceleration?

Physics

1 answer:

PilotLPTM [1.2K]1 year ago

4 0

Answer:

Explanation:

Comment

It's actually a deceleration. You should get a negative answer because the ball is slowing down.

Givens

vi = 8.09 m/s

vf = 5.87 m/s

t = 9.17

Formula

a = (vf - vi) /t Substitute the givens into the formula

Solution

a = (5.87 - 8.09)/9.17

Answer

a = - 2.22 m/s^2

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2 years ago

Read 2 more answers

The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e

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Answer:

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Explanation:

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Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

$|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb$

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

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2 years ago