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3 years ago

URGENT Why is understanding motion so important? Be sure to explain in your own words.

Physics

1 answer:

levacccp [35]3 years ago

3 0

Answer:

Motion is one of the important issues of Physics. We know that if we are not moving but still earth is moving. Motion is a bit of the machinist's Department of Physics. Several laws are provided for explaining the theory of motion and evolution in motion. The motion and transformation are defined in terms of strength which is wanted for creating an object in a motion state.

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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

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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.

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<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 -

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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-

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3 years ago

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

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The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

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There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

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But we know that $g=\frac{GM}{R^{2} }$

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Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

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3 years ago

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*URGENT* Why is understanding motion so important? Be sure to explain in your own words.

URGENT Why is understanding motion so important? Be sure to explain in your own words.