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3 years ago

A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?

Physics

1 answer:

Marianna [84]3 years ago

6 0

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

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1. How much force would you have to apply to a 15kg object in order to accelerate it! a 2 m/s?

mihalych1998 [28]

30N

Explanation:

Given parameters:

Mass of object = 15kg

Acceleration = 2m/s

Unknown:

Force = ?

Solution:

Force is given as the product of mass and acceleration:

F = m x a

m is the mass

a is the acceleration

Inputting the parameters:

F = 15 x 2 = 30N

The unit of force is newtons, N .

Learn more:

Force brainly.com/question/10470406

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3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

4 years ago

Work is done on an object when an applied force causes the object to move in the same direction as the force.

Lelu [443]

Answer:yes

Explanation:

Work is done on an object when an applied force causes the object to move in the same direction as the force

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4 years ago

If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is

expeople1 [14]

Answer:

We can use 2 g H = v2^2 - v1^2 or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

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3 years ago

How will the positions of the police car and the truck compare when they have the same speed and why?

sdas [7]

Answer:

Two cars of equal weight and braking ability are travelling along the same road but combined with other factors it could mean the difference between life.

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3 years ago