F=ma
f?
m=1300kg
a=1.07m\s squared
f=1300kg x 1.07=1391N
Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by
So, .... (1)
Now,
Substitute the value of v from equation (1), we get
Let the radius of the alpha particle is r2.
For proton
So, ... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So, ... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get
q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values
So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
To solve the problem, we can use Ohm's law, which states that
where
V is the voltage
I is the intensity of current
R is the resistance
Using the data of our problem, I=4 A and
, we find the voltage:
Answer:
v = 6i + 12j + 4k
Explanation:
Find the magnitude of the direction vector.
√(3² + 6² + 2²) = 7
Normalize the direction vector.
3/7 i + 6/7 j + 2/7 k
Multiply by the magnitude of v.
v = 14 (3/7 i + 6/7 j + 2/7 k)
v = 6i + 12j + 4k