Answer:
The answer is below
Explanation:
We are to check if the statement is true of false. If it is false, we correct the statement.
Solution:
Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.
The x coordinate represent the time and the y coordinate velocity.
5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:
Acceleration = 
This is correct.
6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:
Acceleration = 
This is false.
Therefore the acceleration of graph B is 0 m/s².
Answer:
500 kg
Explanation:
It is given that,
The mass of a open train car, M = 5000 kg
Speed of open train car, V = 22 m/s
A few minutes later, the car’s speed is 20 m/s
We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :
initial momentum = final momentum
Let m is final mass
MV=mv

Water collected = After mass of train - before mass of train
= 5500 - 5000
= 500 kg
So, 500 kg of water has collected in the car.
Answer:
M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b]
ΔR is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
====================================
Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
----
Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N -----
Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
----
1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
=====
Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * ΔR/R[b]