If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law
K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get
F'= 2 F
Thus the force gets doubled.
Answer:2.55 rad/s
Explanation:
Given
Diameter of ride=5 m
radius(r)=2.5 m
Static friction coefficient range=0.60-1
Here Frictional force will balance weight
And limiting frictional force is provided by Centripetal force
weight of object=mg
Equating two
f=mg
Answer:
The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N
(b) the force the light beam exerts is much too small to be felt by the man.
Explanation:
Given;
cross-sectional area of the man, A = 0.500m²
intensity of light, I = 35.5kW/m²
If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;
P = I/c
where;
P is the pressure of the incident light
I is the intensity of the incident light
c is the speed of light
F = PA
where;
F is the force of the incident light on the man
P is the pressure of the incident light on the man
A is the cross-sectional area of the man
F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N
The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N
Therefore, the force the light beam exerts is much too small to be felt by the man.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s