Answer:
F = 10.788 N
Explanation:
Given that,
Charge 1, 
Charge 2, 
Distance between charges, d = 0.1 m
We know that there is a force between charges. It is called electrostatic force. It is given by :

So, the force applied between charges is 10.788 N.
Answer:
T = g μ_s ( M+m )
78.4 N
Explanation:
When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .
Net force on the lower box
= T - mg μ_s = M a ( a is the acceleration created by net force in M )
Considering force on the upper box
mg μ_s = ma
a = g μ_s
Put this value of a in the equation above
T - m gμ_s = M g μ_s
T = mg μ_s + M g μ_s
= g μ_s ( M+m )
2 )
Largest tension required
T = 9.8 x .50 x ( 10+6 )
= 78.4 N
Answer:
mass of the person walking to west is 65 kg.
Given:
Momentum = 52 
Speed = 0.8 
To find:
Mass of the person = ?
Formula used:
Momentum is given by,
P = m × v
Where, P = momentum
m = mass
v = speed
Solution:
Momentum is given by,
P = m × v
Where, P = momentum
m = mass
v = speed
Mass = 
m = 
m = 65 kg
Thus, mass of the person walking to west is 65 kg.
Answer:
Part A : E =
ε₀ Q₁/R₁² Volt/meter
Part B : V =
ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
<em>Part A</em>
The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F =
ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E =
ε₀ Q₁/R₁² Volt/meter
Part B
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V =
ε₀ Q₁/R₁ Volt
Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)