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
Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire,
=
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, 
where,
=Avogadro's Number

Now,
the drift velocity, 

where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ 

The time will be:
⇒ 



hence,
⇒ 

Upstream speed = S - 1
Downstream speed = S + 1
Average speed = total distance / total time
Average speed = (S - 1) + (S + 1) / 2
= S
S = 6 miles / 4 hours
S = 1.5 miles per hour
Answer:
Final velocity of the car will be -9.28 m/sec
Explanation:
We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec
Acceleration of the car
in negative direction so acceleration will be 
From first equation of motion we know that
v = u+at
So 
So final velocity will be -9.28 m/sec