Given:
B(Magnetic field): 1.5 T
q= 7.5 microcoulombs
v= 1.75 x 10 ∧6 m/s
The angle ∅ between B and v is 45 °.
Now we know that F= qvB sin ∅
Substituting these values we get:
F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45
F= 16.752 N
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
Answer:
The value is 
Explanation:
From the question we are told that
The mass is 
The needed oscillation time is 
Generally the spring constant is mathematically represented as
=> 
=>
Answer:the will have the same acceleration
Explanation:according to fállelos theory they land at the same time
Answer:
0.70046 m/s²
2.732862 N
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of yo-yo = 0.3 kg
R = Radius of rolling disk = 5.1 cm
r = Radius of rod = 1 cm
For a rolling disks the acceleration is given by
The acceleration of the yo-yo is 0.70046 m/s²
The tension in the string will be
The tension in the string is 2.732862 N
