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PolarNik [594]
2 years ago
12

An evacuated tube uses an accelerating voltage of 55 kV to accelerate electrons to hit a copper plate and produce x rays. Non-re

lativistically, what would be the maximum speed of these electrons?
Physics
1 answer:
AleksAgata [21]2 years ago
3 0

Answer:

v = 4.4 x 10⁷ m/s

Explanation:

The kinetic energy of the electrons will be equal to the energy supplied by the electric voltage:

Kinetic Energy = Electric Energy

\frac{1}{2}mv^2 = eV

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

e = charge on electron = 1.6 x 10⁻¹⁹ C

V =Voltage = 55 kV = 55000 V

Therefore,

\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(v)^2 = (1.6\ x\ 10^{-19}\ C)(55000\ V)\\\\v^2 = \frac{(2)(8.8\ x\ 10^{-16}\ J)}{9.1\ x\ 10^{-31}\ kg}\\\\v = \sqrt{19.34\ x\ 10^{14}\ m^2/s^2}

<u>v = 4.4 x 10⁷ m/s</u>

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An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds
Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s

So, the average speed of the object is 6 m/s.

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3 years ago
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

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