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Harman [31]
3 years ago
9

A hiker shouts toward a vertical cliff 685 m away. The echo is

Physics
1 answer:
miv72 [106K]3 years ago
7 0
To find the speed of sound you take the distance and divide it by the time so 685/4 which equals 171.25 m/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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6 0
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What items can be classified as matter?
natita [175]
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4 0
3 years ago
Have a pH between 0-7.
Pavlova-9 [17]

Answer:

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Explanation:

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A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the
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Answer:

refer to the above attachment

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