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oee [108]
3 years ago
13

A sanding disk with rotational inertia 1.2 × 10−3 kg ⋅ m2 is attached to an electric drill whose motor delivers a torque of magn

itude 16 N ⋅ m about the central axis of the disk. About that axis and with the torque applied for 33 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?
Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

L = 0.528kgm²/s

ω = 440 s^-1

Explanation:

given

Inertia, I = 1.2 * 10^-3kgm²

Torque, τ = 16Nm

Time, t = 33ms

Angular momentum is gotten using the relation

τ = ΔL/ΔT

Where τ = Torque

ΔL = change in angular momentum

ΔT = change in time taken

ΔL = τΔT

ΔL = 16Nm * 33*10^-3s

ΔL = 0.528kgm²/s

Angular velocity of the disk will be gotten by using the formula

L = Iω where

L = angular momentum

I = rotational inertia

ω = angular velocity

ω = L/I

ω = 0.538/1.2*10^-3

ω = 440 s^-1

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meriva

Answer:

Explanation:

Given

acceleration of system a =1.2 m/s^2

Normal Force N=4.45 N

Force exerted F=20 N

Thus

F=(m_1+m_2)a

\frac{20}{1.2}=m_1+m_2

16.67=m_1+m_2-------1

Normal reaction N=m_2a

4.45=m_2\times 1.2

m_2=3.70 kg

therefore m_1=16.67-3.70

m_1=12.96 kg

8 0
3 years ago
Read 2 more answers
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
2 years ago
*PLEASE HELP! FIRST CORRECT ANSWER GETS BRAINLIEST!*
scoray [572]

Answer:

Explanation:

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When cloning, the object would be able to create tissue and organs that doctor can use when needed during the surgery.

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6 0
2 years ago
A 1.5 kg ball has a velocity of 12 m/s just before it strikes the floor. Find the impulse on the ball if the ball bounces up wit
Andru [333]

Hi there!

Recall:

Impulse = Change in momentum

I = Δp = mΔv = m(vf - vi)

Let the direction TOWARDS the floor be POSITIVE, and AWAY be NEGATIVE.

Plug in the givan values:

Δp = 1.5(-10 - 12) = -33 Ns

**OR, the magnitude: |-33| = 33 Ns

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2 years ago
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Answer:

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