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Wewaii [24]
3 years ago
9

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

ssume that the platform is initially at rest and the torque is applied for 12.0 rotations. Neglect friction.
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process?  Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

W=\Delta K

W= 1/2*I*\omega f^2-1/2*I*\omega o^2

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

\omega f^2=\omega o^2+2*\alpha*\theta

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

\tau=I*\alpha

\alpha=\tau/I     =>     \alpha = 0.5rad/s^2

Now we can calculate the final velocity:

\omega f = 8.68rad/s

Finally, we calculate the total work:

W= 1/2*I*\omega f^2 = 1883.56J

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

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A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The
qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0
3 years ago
A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.
Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

F=mg

\dfrac{mv^2}{r}=mg

\dfrac{v^2}{r}=g

v=\sqrt{rg}

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

v=\sqrt{250\times9.8}

v=49.49\ m/s

Hence, The speed of space station floor is 49.49 m/s.

6 0
3 years ago
A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. what is the volume of the ball?
lesantik [10]
Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
6 0
3 years ago
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How much current is in a circult that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms?
SVETLANKA909090 [29]

Answer:

Current, I = 2.3 A

Explanation:

We have,

Voltage of the battery in a circuit is 9 volts

Resistance of the circuit is 4 ohms

It is required to find the current in a circuit. When the voltage and the resistance of the circuit is given then we can find the current in it is given by Ohm's law as :

V=IR

I is electric current

I=\dfrac{V}{R}\\\\I=\dfrac{9}{4}\\\\I=2.25\ A

or

I = 2.3 A

So, the current in the circuit is 2.3 A.

6 0
3 years ago
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Photovoltaic cells are the most efficient means of converting solar energy to electricity. Option b is correct.

<h3>What is a cell?</h3>

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One or more cells can make up a battery. One cell, for example, is one AA battery.

Light intensity on a solar cell is often measured in "suns," with one sun roughly equivalent to 1 kW/m².

Concentrated sunlight improves the ratio of current generated while the device is lighted vs when it is dark, hence enhancing output voltage and efficiency.

Photovoltaic cells are the most efficient means of converting solar energy to electricity.

Hence, option b is correct.

To learn more about the cell refer to:

brainly.com/question/3142913

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8 0
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