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lys-0071 [83]
3 years ago
15

Can someone help with these questions?

Physics
1 answer:
yanalaym [24]3 years ago
6 0

One

Givens

  • v = 8.5 km/s = 8500 m/s
  • R = 25 cm = 0.25 m
  • B = 5.29 * 10^-3
  • q = 1.6*10^-19
  • m = ??

Formula

m*v^2/ R = qvB    Cancel 1 v on each side.

m*v / R = q*B

which for m when transposed is

m = q*B*R/v

m = 1.6*10^-19 * 5.29*10^-3 * 0.25 / 8500

m = 2.49 * 10^-26

Answer: A

Two

Givens

  • B = 0.5 T
  • q = 1.6 * 10^ -19
  • v = ??
  • m = 3.34 * 10^-27
  • R = 55.6 cm = 0.556 m

Formula

qvB = mv^2/R    Cancel one of the "v"s

qB = mv/R

qBR / m = v

Solution

1.6*10^-19 * 0.5 T * 0.556 m / 3.34 * 10^-27

13317365 = v

1.33 * 10^7 = v

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sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

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now we know that

object distance from mirror is

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Focal length of the mirror is given as

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now we have

\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}

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6 0
3 years ago
You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

= 2.64 m/s

Explanation:

Working with the forces acting on the water in the pail at any point.

The weight of water is always directed downwards.

The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

Net force = W

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Hope this Helps!!!

7 0
3 years ago
A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

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(a)ω = 1 rad/s

(b)t = 2.41 s

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0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

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b.)

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As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

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