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Goryan [66]
3 years ago
5

You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1

.85 m to 1.50 m. What is its change in potential energy?
Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

Change in potential energy = 8.58 J

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

Potential energy = mgh

m = 2.50 kg

g = 9.81 m/s²

Initial height, h₁ = 1.85 m

Final height, h₂ = 1.50 m

Change in potential energy = mgh₁  - mgh₂

Change in potential energy = 2.5 x 9.81 x 1.85 - 2.5 x 9.81 x 1.50

Change in potential energy = 8.58 J

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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
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Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

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3 years ago
A 1,200 kg dragster, starting from rest, reaches a maximum velocity of 140m/s in 5 seconds. At the 5 second mark, the dragster d
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Drag or air resistance

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