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2 years ago

How the forces makes effect on the state of motion of an object ? Plz ans fast urgent

Physics

1 answer:

Ksju [112]2 years ago

6 0

Answer:

1. It may change the direction of an object in motion.

2. It may cause change in velocity of an object in motion.

Explanation:

1.It may change the direction of an object in motion.

When an object is in motion,an applied force on that object may change its direction.

For example, a sailboat moving eastward, can suddenly change its direction by interaction of a storm wind blowing form the south.

2. It may cause change in velocity of an object in motion .

A force applied to an object in motion can increase or decrease its speed. When the force is applied to the object in motion in the direction of that object, its velocity may increase.

On the other hand, when the force is applied in the opposite direction to the object in motion, its velocity may reduce.

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A dart hits a dartboard and stops in 0.060 s. The net force on the dart is 14 N during the collision.

Rainbow [258]

Answer:

<em>The change of momentum of the dart is 0.84 Nw.s</em>

Explanation:

<u>Impulse and change of momentum</u>

The change in momentum of an object is its mass times the change in its velocity:

$\Delta p=m\Delta v=m(v_2-v_1)$

The change in the momentum can also be found by considering the force acting on it. If a force F acts for a time Δt, the change of momentum is given by:

$\Delta p=F.\Delta t$

The dart hits a dashboard with a net force of 14 N during the collision and stops in 0.06 seconds. The change of momentum is:

$\Delta p=14*0.06=0.84$

The change of momentum of the dart is 0.84 Nw.s

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

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