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taurus [48]
3 years ago
14

A wheel with rotational inertia i is mounted on a fixed, frictionless axle. the angular speed ω of the wheel is increased from z

ero to ωf in a time interval t. what is the average net torque τ on the wheel during this time interval?
Physics
1 answer:
Annette [7]3 years ago
4 0
According to Newton's second law:
τ = Iα, where I is the moment of inertia and α is the angular acceleration.
Angular acceleration can be written as:
α = (ωf - ωi) / t
α = ωf/t

Substituting into the second law:
τ = Iωf/t
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zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

L_{i}=L_{f}

Here,  = the system's angular momentum before the collision

L_{i} = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here,  = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

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4 0
1 year ago
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can I skip gravitation and tissue chapter for class 9 annual examination?? which lessons are most important and which aren't??
velikii [3]
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8 0
3 years ago
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with
Dvinal [7]

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

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6 0
3 years ago
A ball is dropped from some height. It bounces off the floor and rebounds with a speed that is one-half the speed it had just be
Arada [10]

Answer:

The correct option is C

Explanation:

According to third equation of motion, v

2

=u

2

+2ax

Here, u=0 m/s

a=−g and x=−h

Negative sign indicates downward direction. Displacement and acceleration both are downwards.

So,v=±

2(−g)(−h)

​

We take minus sign because it is downwards.

v=−

2gh

​

After bouncing. velocity becomes 80% of v, i.e.,

v

′

=+0.8

2gh

​

 

(positive sign because the direction of ball has reversed after bouncing and is upwards.

Applying third equation of motion again, for u=v

′

, v=0 and a=−g

v

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Thus,

0=0.64(2gh)+2(−g)x

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x=0.64h

3 0
2 years ago
15. When you cannot stop safely at a yellow traffic light before entering an intersection, ______________. A. stop in the inters
Elan Coil [88]

Answer:  When you cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Explanation: To find the correct answer, we need to know more about the traffic signal rules.

<h3>What is the traffic signal rules?</h3>
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Thus, we can conclude that, the option C is correct.

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