The answer is 68 F. i hope this helps
Answer:0.318 revolutions
Explanation:
Given
Initially Propeller is at rest i.e. 
after 

using 


Revolutions turned in 2 s



To get revolution 
=
Copper because it contains alot of electricity
Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²