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3 years ago

8

A photon of wavelength 0.99235 nm strikes a free electron that is initially at rest. the photon is scattered straight backward.

what is the speed of the recoil electron after the collision?

1 answer:

Tomtit [17]3 years ago

5 0

F=uR
where U is constant of friction and R=mg
F= mgcos 36
=2.05*10 *.8090
=16.58 N

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The radius of Saturn is about 10 times the radius of Venus and the mass is about 100 times that of Venus. How much larger is the

lora16 [44]

let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

$F_1 = \frac{GMm}{R^2}$

now similarly the gravitational force on the man if he is at the surface of Saturn

$F_2 = \frac{G*100M*m}{(10R)^2}$

$F_2 = \frac{GMm}{R^2}$

so here if we divide the two forces

$\frac{F_1}{F_2} = 1$

so here we can say

F1 = F2

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3 years ago

A straight wire of length 4.5 cm moves at a constant speed of 5.2m/s perpendicular to its length and a uniform magnetic field. I

liraira [26]

Well you have to minus the 4.5 to 5.2 and the answer to that would be -11.5 and calculated that to be 4.5

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3 years ago

How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.

guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R <span>

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>

<span>and since: ω = 2pi*f --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
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5 0

3 years ago

800 joules of work were done with a force of 200 newtons. Over what

Svetach [21]

Answer:

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Explanation:

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3 years ago

How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume

Svetach [21]

Answer:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$

7 0

3 years ago