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3 years ago

If a 60 kg student is standing on the edge of a cliff. Find the students gravitational potential of the cliff is 30 m high

Physics

1 answer:

kipiarov [429]3 years ago

8 0

Answer:

17640J

Explanation:

Give mass m= 60kg

Height h =30m

Since the student is standing on the edge of a cliff , acceleration due to gravity g is 9.8m/s^2

The student’s gravitational potential energy can be found by the formula

mgh

That’s 60 x 9.8 x 30

= 17640J

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A uniform drawbridge must be held at a 37 ? angle above the horizontal to allow ships to pass underneath. The drawbridge weighs

iVinArrow [24]

Answer:

Explanation:

a ) Let T be tension in the horizontal cable .

Taking torque about hinge of weight and tension

45000 x 7 x cos37 = 3.5 sin 37 x T

T = 119431 N

b ) If R be the reaction force on the hinge

R sin 37 = 45000 ( vertical force balancing each other )

R = 74775 N

c ) The direction of R will be along the bridge

d ) Moment of inertia of bridge

I = Ml² / 3 = (45000 / 9.8 ) x 14² / 3

= 300000.

When cable breaks , the weight creates a torque about the hinge will creates angular acceleration

Torque by weight about the hinge

= 45000 x 7 x cos 37

= 251570

angular acceleration = torque / moment of inertia

= 251570 / 300000

= .84 radian / s²

4 0

3 years ago

At Alpha Centauri's surface, the gravitational force between Alpha Centauri and a 2 kg mass of hot gas has a magnitude of 618.2

sergeinik [125]

Answer:

6.86 * 10^8 m

Explanation:

Parameters given:

Mass of hot gas, m = 2 kg

Gravitational Force, F = 618.2 N

Mass of Alpha Centauri, M = 2.178 * 10^30 kg

The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:

F = (G*M*m) / r²

Where G = gravitational constant

Therefore,

618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²

=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2

r² = 4.699 * 10^17 m²

=> r = 6.86 * 10^8 m

We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.

The mean radius of Alpha Centauri is 6.86 * 10^8 m.

5 0

3 years ago

Read 2 more answers

a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)

rodikova [14]

The x-component of the normal force is equal to 1706.45 N.

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

$N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)$

Now, using the given information, we have:

$mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}$

Calculating, we have:

$N_{x}=mg*Sin(\alpha)$

$N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N$

Hence, we have that the x-component of the normal force is equal to 1706.45 N.

Have a nice day!

3 0

4 years ago

Two compounds are standing at the same temperature. Compound "A" is evaporating more slowly than compound "B." According to the

Whitepunk [10]

I think that some options might have helped to answer the question. I am answering the question based on my research and knowledge. "Compound B may have a lower molecular weight" is the assumption that can be made according to the Kinetic Molecular Theory. I hope the answer has come to your help.

7 0

3 years ago

Read 2 more answers

A 25-turn circular coil of wire has diameter 1.00 m. It is placed with its axis along the direction of the Earth’s magnetic fiel

slava [35]

Answer:

The magnitude of average emf is 9.813 mV.

Explanation:

Given that,

Number of turns = 25 turns

Diameter = 1.00 m

Magnetic field = 50.0μm