I believe the answer is b) slowly heating the surface
Answer:
Using the given values
F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N
E = 9 * 1.6^2 / 5.18^2 * 10 = 8.5 N
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8
Explanation:
in the attachment
Answer:
The magnitude of angular acceleration is
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is
. Hence, this is the required solution.