Answer:
<em>Speed of the electron is 2.46 x 10^8 m/s</em>
<em></em>
Explanation:
momentum of the electron before relativistic effect = 
where 
is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M = 
where
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==> 
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75 =
the equation reduces to
1.75 = 
square both sides of the equation, we have
3.0625 = 1/
3.0625 - 3.0625
= 1
2.0625 = 3.0625
= 0.67
β = 0.819
substitute for
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>
Answer:
0.3817 N
Explanation:
Remark
One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.
Givens
m = 101 g
vi = 0 That's what at rest means.
t = 2.91 s
d = 16 m
F= ?
Formulas
d = vi*t + 1/2*a * t^2
Force = m * a
Solution
16 = 0 + 1/2 a * 2.91^2
16 = 4.234 a Divide by 4.234
16/4.234 = a
a = 3.779
F = m * a
a = 3.779
m = 101 g = 1 kg / 1000 grams
m = 0.101 kg
F = 0.101 * 3.779
F = 0.3817N
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.
for 1 degree................... 7 Joules
y given degree........ p Joules
p=7y
In our case y=(t-25) .
h(t) = 7(t-25) which is the final answer.
Answer:
i think number 2 should be your pfp
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
is the coefficient of friction, while N is the normal force. So we have:
since we know that F=300 N and
, we can find N, the magnitude of the normal force:
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
due to the presence of the oil. Therefore, we have:
