Answer: The electrostatic force will be the same
Explanation:
According to Coulomb's Law, when two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies or particles.
In this sense, this law states the following:
"The electrostatic force between two point charges and is proportional to the product of the charges and inversely proportional to the square of the distance that separates them, and has the direction of the line that joins them"
(1)
Being is a proportionality constant.
Now, if each and are increased by 6, and the distance between them as well, we will have the following:
(2)
(3)
Simplifying:
(4)
Comparing (1) with (4) we can see the electrostatic force is the same.
The car is gradually increasing in distance as time goes on
Answer:
<em>45.375m</em>
Explanation:
To get the distance travelled by the body we will use the equation of motion
S = ut + 1/2at²
u is the initial speed = 0m/s
a is the acceleration
t is the time = 5 seconds
We need to get the acceleration first:
According to Newton's second law,
Fm is the moving force = Wsin theta
Ff is the frictional force = 1500N
m is the mass = W/g
m = 4000/9.8
m = 408.16kg
Substitute the given values in the formula and get the acceleration:
Get the required distance:
Recall that S = ut+1/2at²
S = 0(5) + 1/2*3.63*5²
S = 0+0.5*3.63*25
S = 45.375m
<em>Hence the distance travelled by the body in 5 seconds if the force of friction is 1500N is 45.375m</em>
Answer:
Explanation:
<u>Work
</u>
The work done by an external force F when moving an object by a distance X parallel to its direction is
The second Newton's law gives us the net force acting on an object of mass m which is being moved at an acceleration a:
Replacing the last equation into the first equation
And solving for a
There seems to be some kind of mistake in the data provided, we'll use it anyway
It's a very small value as compared to the dimensions of the rest of the problem.
Answer:
Explanation:
a )
Magnetic field inside solenoid B = μ₀ NI ,
μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B= 4π x 10⁻⁷ x 30 x 10² x 15
= .0565 T .
Force on each side of square loop = B i L
B is external magnetic field , i is current in loop and L is length of side
Force on each side of square loop = .0565 x .24 x 2 x 10⁻²
= 2.7 x 10⁻⁴ N .
b )
Torque on the loop = F x d
F is force on one side , d is distance between two sides , that is side of the square loop
= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m
= 5.4 x 10⁻⁶ N.m .