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ss7ja [257]
4 years ago
11

An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston

e is I = 2.0 kgm2 . The system starts from rest. A. Determine the angular acceleration of the shaft when this torque is applied. B. Determine the kinetic energy of the shaft after 8 seconds of operation. C. Determine the work done by the motor in this time. D. Determine the average power delivered by the motor over this time interval
Physics
1 answer:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

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A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

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

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
3 years ago
A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth
Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

W=mg

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center


The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

800 N= W=mg=m \frac{GM}{R^2} (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

200 N = W' = mg' = m \frac{GM}{(R+h)^2} (2)

If we divide eq.(1) by eq.(2), we get

\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}

4=\frac{(R+h)^2}{R^2}

By solving the equation, we find:

4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0

which has two solutions:

h=-3R --> negative solution, we can ignore it

h=R --> this is our solution

Since the Earth's radius is R=6.4\cdot 10^6 m, the person should be at h=R=6.4\cdot 10^6 m=6400 km above Earth's surface.

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