On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
Answer:
32 m and -2.4 m/s
Explanation:
Given:
v₀ = 25 m/s
t = 2.8 s
a = -9.8 m/s²
Find: Δy, v
Δy = v₀ t + ½ at²
Δy = (25 m/s) (2.8 s) + ½ (-9.8 m/s²) (2.8 s)²
Δy = 31.6 m
v = at + v₀
v = (-9.8 m/s²) (2.8 s) + 25 m/s
v = -2.44 m/s
Rounded to two significant figures, the bullet reaches a height of 32 m and a velocity of -2.4 m/s.
Answer:
Zero.
Explanation:
An adiabatic process is one in which there is no exchange of heat energy. Therefore, in an adiabatic process, heat is neither added to the system not it is removed from the system.
The work done by the gas on the environment is 20 J. This energy is equal to the change in internal energy for an adiabatic process.
Therefore, for an ideal gas to undergo an adiabatic process in which it expands and does 20 J of work on its environment, the heat exchange is zero.
The speed of light : 299 792 458 m / s
