It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.
Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.
Answer:
P(bat) = V²r/(R+r)²
Explanation:
Let the resistance of the coil be R
Internal resistance of the battery be r
Emf of the battery = V
Power dissipated in the internal resistance of the battery is normally given as P = I²r
where I is the current flowing in the circuit.
From Ohm's law,
V = I R(eq)
R(eq) = (R + r)
I = V/(R+r)
P = I²r
P = [V/(R+r)]²r
P = V²r/(R+r)²
Hope this Helps!!!
The magnitude of the electrostatic force between two charges is given by:

where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s