I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
The law of conservation of matter states that matter cannot be created nor destroyed. It can only be transformed from one form to another. To state an example where this is shown, let's say a piece of paper is burning. Not having a scientific background, you would say that the matter is being destroyed. But in reality, the paper is simple being transformed to ash, carbon dioxide, and water vapor. Overall, the total mass would still remain the same.
Answer:
In most materials, as heat energy is absorbed, the density decreases. IF a certain object is heated, it might appear bigger than usual because it expands as the molecules inside moves faster than usual. However the mass of it stays the same while the density decreases.
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:
t = 0.354 hours
Explanation:
given,
coefficient of rolling friction μr=0.002
mass of locomotive = 180,000 Kg
rolling speed = 25 m/s
The force of friction = μ mg
= (.002) x (180000) x (9.8)
= 3528 N
F = m a
now,
m a = 3528 N
180000 x a = 3528
a = 0.0196 m/s²
Then apply
v = u + at
0 = 25 - 0.0196 x t
t = 1275.51 sec
t = 1275.61/3600 hours
t = 0.354 hours
time taken by the locomotive to stop = t = 0.354 hours
