An object travels along a straight line across a horizontal surface, and its motion is described by the velocity versus time gra
1 answer:
Answer:
C.)Finding the area bound by the horizontal axis and the curve from 0 s to 5 s
D.)Using Average Speed = total distance/total and multiplying the average speed of 5m/s by a total time of 5 s
Explanation:
The graph seems to be missing, however we can still answer the question.
In fact, we have to keep in mind the following facts:
- In a velocity-time graph, the area under the curve represents the displacement of the object, while the slope of the curve corresponds to its acceleration.
From this, we can already say that the following option:
C.)Finding the area bound by the horizontal axis and the curve from 0 s to 5 s
is correct, while the statement
A.)Finding the slope of the line between 0 s to 5 s
is wrong.
Moreover, the average speed of an object in motion is given by
where d is the total distance covered while t is the total time taken. Once we have calculated the average speed, the total displacement over a certain time period can be simply calculated by multiplying the average speed by the time interval: so, the option
D.)Using Average Speed = total distance/total and multiplying the average speed of 5m/s by a total time of 5 s
is also correct.
The last option remained:
B.)Dividing the change in velocity between 0 s to 5 s by the change in time
is wrong, because in this way we are calculating the acceleration, not the displacement.
You might be interested in
Answer:
The charge on the capacitor had increased
Explanation:
The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;
Here, C is the capacitance, is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;
Here, K is the dielectric constant.
In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.
The expression for the charge stored in the capacitor is as;
Q= CV
Here, Q is the charge and V is the potential.
The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.
The expression for the energy stored in the capacitor is as follows;
The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.
The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.
Therefore, only option (a) is true.