Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Explanation:
Solution,
- Mass(m)= 60 kg
- Force (F)= 20 N
- Acceleration (a)= ?
We know that,
- F=ma
- a=F/m
- a=20/60
- a=0.333 m/s²
So, her acceleration is 0.333 m/s².
Answer:
The forces that do non-zero work on the block are gravity and normal reaction force
Explanation:
Answer: F
Out of the page.
Explanation:
For an electron with a charge of -e, the magnitude of the force on it is F = BeV
Where
F = force on the electron
e = charge ( electrons )
V = velocity
B = magnetic field
F is the force acting on all the electrons in a wire which gives rise to the F = BIL
Where
I = current
L = length of the wire
The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.
