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MAVERICK [17]
3 years ago
15

A forklift raises a 72-kg crate onto a shelf 3.6 meters above the floor. How much gravitational potential energy does the crate

have relative to the floor?​
Physics
1 answer:
Mademuasel [1]3 years ago
8 0

Answer:

<h2>2540.16 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 72 × 9.8 × 3.6 = 2540.16

We have the final answer as

<h3>2540.16 J</h3>

Hope this helps you

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-4, 0, -2/3, 4.11111…, 2, π, √6 'which members of this set are irrartional?
krok68 [10]

An irrational number is a real number that cannot be written as a simple fraction.

4 = 4/1 = rational

0 = 0/1 = rational

-2/3 = rational

4.11111 = irrational

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√6= irrational

Answer:

4.1111, π , √6

3 0
1 year ago
b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s
fomenos

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m

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Which of the following is the best definition of an isotope?
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4 0
3 years ago
If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0
3 years ago
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