The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.
<h3>What is acceleration?</h3>
Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.
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Answer: 2000 watts
Explanation:
Given that,
power = ?
Weight of object = 200-N
height = 4 m
Time = 4 s
Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.
i.e Power = work / time
(since work = force x distance, and weight is the force acting on the object due to gravity)
Then, Power = (weight x distance) / time
Power = (200N x 4m) / 4s
Power = 8000Nm / 4s
Power = 2000 watts
Thus, 2000 watts of power is needed to lift the object.
<h2>Given that,</h2>
Mass of two bumper cars, m₁ = m₂ = 125 kg
Initial speed of car X is, u₁ = 10 m/s
Initial speed of car Z is, u₂ = -12 m/s
Final speed of car Z, v₂ = 10 m/s
We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

v₁ is the final speed of car X.

So, car X will move with a velocity of -12 m/s.
Answer:

Explanation:
Given data

To find
Mutual inductance of the two-coil system
Solution
The mutual inductance given as:
M= (-VΔt)/ΔI
Substitute the given values
So

Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
