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2 years ago

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Ben starts walking along a path at 3 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the s

ame path at 6 6 mi/h. How long will it be before Amanda catches up to Ben?

1 answer:

Drupady [299]2 years ago

4 0

Answer:

3 hours

Explanation:

Given:

- The speed of Ben v_b = 3 mi/h

- The speed of Amanda v_a = 6 mi/h

- The total time taken to cover distance(d) by ben = t_b

Find:

How long will it be before Amanda catches up to Ben?

Solution:

- The distance d traveled by Ben:

d = v_b*t_b

d = 3*t_b

- The distance d traveled by Amanda:

d = v_a*t_a

d = 6*t_a

- Equate the distance as when they meet:

3*t_b = 6*t_a

- Where ,

t_b = t_a + 1.5

t_a = t_b - 1.5

- Substitute the time relationship in distance relationship:

3*t_b = 6*(t_b - 1.5)

3*t_b = 6*1.5

t_b = 2*1.5 = 3 h

- Hence, It would take 3 hours since Ben starts walking that amanda catches up.

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<u>Answer:</u>

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<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

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Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

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$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

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A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?

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Given, $S = 20 \ ft$ $S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft} = 6.096 \ m$

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