6 is b. part B on 6 is a. 7 is a. partB ON 7 b
There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.
Answer:
45000 K .
Explanation:
Given :
A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure
We need to find the temperature in which 1 litre of the same gas weigh 1 gram
in pressure 75 atm.
We know, by ideal gas equation :

Here , n is no of moles , 
Putting initial and final values and dividing them :


Hence , this is the required solution.
Explanation:
Given that,
Initial speed of the bus, u = 0
Acceleration of the bus, a = 0.5 m/s²
Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.
So,

Let d is the distance cover during that time. So,

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is,
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as
, and the mass of the first and second coals as
and
respectively
We start with the transition between parts A and B, so we have:

Which means

And since we want the mass of the first coal thrown (
) we do:



Substituting values we obtain

For the transition between parts B and C, we can write:

Which means

Since we want the new final speed of the car (
) we do:

Substituting values we obtain
