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anygoal [31]
3 years ago
14

A cat weighing 5kg is climbing at the top of a tree and has potential energy at 1176kg. Find the height of the tree

Physics
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

23.9m

Explanation:

step one:

This problem is on energy.

we can proceed by using the expression for potential energy to solve for height h

we know that

PE=mgh

given that

mass m= 5kg

PE= 1176 joules

step two:

we can substitute our data and find h

1176= 5*9.81*h

1176=49.05h

step three:

divide both sides by 49.05 we have

1176/49.05=h

h=23.9 m

The height of the tree is 23.9m

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How long does it take the lava bomb to reach its maximum height? Answer with three significant digits and the correct unit. A sm
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Answer:

The time taken to reach the maximum height is 3.20 seconds

Explanation:

The given parameters are;

The initial height from which the volcano erupts the lava bomb = 64.4 m

The initial upward velocity of the lava bomb = 31.4 m/s

The acceleration due to gravity, g = 9.8 m/s²

The time it takes the lava bomb to reach its maximum height, t, is given by the following kinematic equation as follows;

v = u - g·t

Where;

v = The final velocity  = 0 m/s at maximum height

u = The initial velocity = 31.4 m/s

g = The acceleration due to gravity = 9.8 m/s²

t = The time taken to reach the maximum height

Substituting the values gives;

0 = 31.4 - 9.8 × t

∴ 31.4 = 9.8 × t

t = 31.4/9.8  ≈ 3.204

The time taken to reach the maximum height rounded to three significant figures = t ≈ 3.20 seconds

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3 years ago
How long would it take a person to come to a stop if they were going
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Which of the following is a result of gravitational forces in the Solar System?
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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
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Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

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