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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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Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

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We are told the values of \Phi and B, then we can calculate the magnitude

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3 0
3 years ago
NEED HELP ASAP!!!!!!!!!!!!!
love history [14]

Answer: Decreased risk of heart attack

Explanation:

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Hope this helps :)

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An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
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Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

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F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

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