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storchak [24]
3 years ago
13

A dad takes his kids to their school just 8.0 miles down the road but with traffic it takes him 30 minutes and the fastest he ca

n go is 35 mi/hr. One of his children has forgotten his lunch, so the dad goes back, gets the lunch, and returns to the school. It takes him just an additional 30 minutes since traffic has cleared. What is his average velocity for the morning?
A 8 mi/hr
B 16 mi/hr
C 24 mi/hr
D 35 mi/hr
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:C 24 mi/hr

Explanation:

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If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m
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Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

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So we can write

mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}

We can cancel the common factor of m which leaves us with

gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}

Lets solve for v_f

gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}

Subtract gh_f from both sides of the equation.

gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}

Multiply both sides of the equation by 2.

2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}

Simplify the left side.

Apply the distributive property.

2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}

Cancel the common factor of 2.

2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}

Take the square root of both sides of the equation to eliminate the exponent on the right side.

{v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}

We are given g,v_{i},h_{i},h_{f}.

We can now solve for the final velocity.

{v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)

Anything multiplied by 0 is 0.

{v_{f}=\sqrt{2*9.81*2.41

{v_{f}=\sqrt{47.2842

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