The answer would be East. Negative particle. if that reads west then the opposite is east.
The planets near the sun are smaller (and made out of rock or earth )
Answer:
The magnitude of the field is 8.384×10^-4 T.
Explanation:
Now, i start solving this question:
First, convert the potential difference(V) 2 kv to 2000 v.
As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so
r = 0.36/2 = 0.18 m.
As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.
We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.
qV = 1/2mv^2
(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2
v = 2.65×10^7 m/s.
These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:
qvB = mv^2/r
qB = mv/r
B = mv/qr
B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)
Hence, B = 8.384*10^-4 T.
Answer:
A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.
Gpe @ A=mgh=40*9.81*2=784.8J
B. At 30degree vertical angle the vertical displacement from lowest position is given by
2-2cos(30)=2-1.73=0.27m
Gpe @B= 40*9.81*0.27=106 J
C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.
The energy the electron gives up passing between the electrodes is equal to the product between its charge and the potential difference between the electrodes:
where
e is the electron charge
is the potential difference
Plugging numbers into the equation, we find that the electron gives up is
