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.
Answer:
Option C. 30 m
Explanation:
From the graph given in the question above,
At t = 1 s,
The displacement of the car is 10 m
At t = 4 s
The displacement of the car is 40 m
Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:
Displacement at t = 1 s (d1) = 10 m
Displacement at t= 4 s (d2) = 40
Displacement between t = 1 and t = 4 (ΔD) =?
ΔD = d2 – d1
ΔD = 40 – 10
ΔD = 30 m.
Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.
Answer:
Explanation:
Time taken to complete one revolution is called time period.
So, Time period, T = 1 s
Diameter = 1.6 mm
radius, r = 0.8 mm
Let the angular speed is ω.
The relation between angular velocity and the time period is

ω = 2 x 3.14 = 6.28 rad/s
The relation between the linear velocity and the angular velocity is
v = r x ω
v = 0.8 x 10^-3 x 6.28
v = 0.005 m/s
Given:
Circumference = 2 m
Angular speed, ω = 1 rev/s = 2π radians/s
If the radius is r, then
2πr = 2
r = 1/π m
The linear (tangential) speed is
v = rω
= (1/π m)*(2π rad/s) = 0.5 m/s
Answer: 0.5 m/s