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vekshin1
2 years ago
11

An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6

kg⋅m2.What is the magnitude of the angular acceleration of the object?
[Please answer if you know]
Physics
1 answer:
Korolek [52]2 years ago
3 0

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law.  Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

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The setting sun is actually visible after it has dropped below the horizon due to the bending of light waves entering the atmosp
Yuki888 [10]
That's one result of "refraction".
4 0
3 years ago
Read 2 more answers
Newton's First Law says that an object at rest will stay at rest, and an object in motion will stay in motion unless:
Verdich [7]

Answer: Acted on by equal forces in opposite direction

Explanation:

Newton's First Law says that every body continue in its state of rest or constant speed on a straight line unless being acted upon by an external force.

6 0
4 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
4 years ago
a circular loop is hanging on the wall. it has a radius of 33.3 cm and is comprised of 12 coils. there is a magnetic field perpe
Vika [28.1K]

Answer:

I = 2.19A, anticlockwise direction.

Explanation:

Given r = 33cm = 0.33m, N = 12, ΔB = 7.5 - 1.5 = 6.0T, Δt = 3s, R = 3.75Ω

By Faraday's law of electromagnetic induction when there is a change in flux in a coil or loop, an emf is induced in the coil or loop which is proportional to the time rate of change of the magnetic flux through the loop.

The emf E is related to the flux by the formula

E = – NdФ/dt

Where N = number of turns in the coil, Ф = magnetic flux through the loop = BA, B = magnetic field strength, A = Area

In this problem the strength of the magnetic field changes. As a result the flux too changes and an emf is induced in the coil.

So

ΔФ = ΔB×A = ΔB×πr² = 6×π×0.33² = 2.05Wb

E = -NΔФ/Δt = 12×2.05/3 = 8.2V

I = E/R = 8.2/ 3.75 = 2.19A

The direction of the current can be found by pointing the thumb of your right hand in the direction of the magnetic field and curling the remaining fingers around this direction. The direction of the curl of these fingers give the direction of current which in this case is anticlockwise.

5 0
4 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

I =0.011 A

5 0
3 years ago
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