The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
The angular momentum(L) of an electron moving in a circular path is given by the formula,
L = mvr ........(i)
We know that the radius of the path of an electron in a magnetic field is
r = mv/qB
Putting this value in equation (i),
L = mv x mv/qB
or L = (mv)^2/qB
Putting the given values in the above equation,
4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3
v comes out to be 8.88 x 10^7 m/s.
Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
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I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity.
The answer is c 1386j
This calculator is very helpful I use it on my homework
https://www.omnicalculator.com/physics/specific-heat
Answer:0.58 m
Explanation:
The initial velocity of the ball is u = 2.0 m/s
The height of the table is, h = 1.0 m
The ball falls in vertical direction under acceleration due to gravity.
Time taken for ball to hit the floor:
h= ut + 0.5gt² ( from the equation of motion)
1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²
Solving this for t,
t = 0.29 s ( we have neglected the negative value of t)
In the same time, the ball would cover a horizontal distance of :
s = u t
⇒s = 2.0 m/s×0.29 s = 0.58 m
Thus, the landing spot is 0.58 m away from the table.
