All categories

3 years ago

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Physics

1 answer:

Virty [35]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has 29,400 Joules of potential energy.

You might be interested in

The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box

Anna [14]

We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

$F_f=\mu N$

Where.

$\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}$

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

$\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}$

Adding the forces in the y-direction we get:

$\Sigma F_y=N-mg_y$

Since there is no movement in the y-direction the sum of forces must be equal to zero:

$N-mg_y=0$

Now we solve for the normal force:

$N=mg_y$

To determine the y-component of the weight we will use the trigonometric function cosine:

$\cos 40=\frac{mg_y}{mg}$

Now we multiply both sides by "mg":

$mg\cos 40=mg_y$

Now we substitute this value in the expression for the normal force:

$N=mg\cos 40$

Now we substitute this in the expression for the friction force:

$F_f=\mu mg\cos 40$

Now we substitute the given values:

$F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40$

Solving the operations:

$F_f=15.01N$

Therefore, the force of friction is 15.01 Newtons.

3 0

1 year ago

PLEASE ANSWER QUICK!! A student pushes a wagon full of bricks with a constant force across the ground. Which of

givi [52]

Answer:

Explanation:

If you do B, the wagon slows down due to friction

If you do C, it would slow down the wagon by half

If you do D, it would slow it down, due to there being more weight but same force.

5 0

3 years ago

A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh we need to know h

h = 50 sin 10 = 8.68 meters

then: PE = 20 * 9.81 * 8.68 = 1703.49 j

KE = 1/2 m v^2 = 1703 .49

v = 13 m/s

7 0

2 years ago

At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit

Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation $a=v^2/r$

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

$v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s$

7 0

3 years ago

Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0

3 years ago