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elena55 [62]
3 years ago
7

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

ould the velocity of a neutral particle be selected by passage through this device?) The explanation of this is that the mass and the charge control the resolution of the device--particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. If the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. Because the acceleration depends on the mass and charge, these influence the sharpness (resolution) of the transmitted particles.
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with __________.

a) both q and m large
b) q large and m small
c) q small and m large
d) both q and m small
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.
babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

p=\frac{F}{A}

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

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Hope it helped you.

My name is Charlie.

Today this question is about science,or chemistry.

A. Acceleration is a correct answer. Acceleration is a rate of change in velocity. 

It change in velocity/change in time. Next, to slope will be indicated by the acceleration.

B Change in position is a incorrect answer.

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-Charlie

Have a great day!

Thanks!
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