I am not completely sure, but I believe that it depends on the total mass of the Protons and Neutrons
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
Answer:
Part a) 
Part b) 
Explanation:
Part a) what is its frequency, in rev/s
we have that
An old-fashioned LP record rotates at 33 1/3 RPM
so

Convert mixed number to an improper fraction

Remember that

Convert rev/min to rev/sec

Simplify

Part b) what is it period, in seconds
we know that
The period is the reciprocal of the frequency
therefore
the frequency is

Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).
Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2
Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.
Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:
acceleration in a circular movement= ω^2*r
so
a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2
a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2