Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
Hope it helped u if yes mark me BRAINLIEST!
Tysm!
:)
Answer:
θ = 28.9
Explanation:
For this exercise let's use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where we use index 1 for air and index 2 for water where the fish is
sin θ₂ = n₁ / n₂ sin θ₁
in this case the air repair index is 1 and the water 1.33
we substitute
sin θ₂ = 1 / 1.33 sin t 40
sin θ = 0.4833
θ = sin⁻¹ 0.4833
θ = 28.9
Solution :
Speed of the air craft, 
= 262 m/s
Fuel burns at the rate of, 
= 3.92 kg/s
Rate at which the engine takes in air, 
= 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft, 
=921 m/s
Therefore, the upward thrust of the jet engine is given by
F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= 
Therefore thrust of the jet engine is .
Answer:
The number density of the gas in container A is twice the number density of the gas in container B.
Explanation:
Here we have
P·V =n·R·T
n = P·V/(RT)
Therefore since V₁ = V₂ and T₁ = T₂
n₁ = P₁V₁/(RT₁)
n₂ = P₂V₂/(RT₂)
P₁ = 4 atm
P₂ = 2 atm
n₁ = 4V₁/(RT₁)
n₂ =2·V₁/(RT₁)
∴ n₁ = 2 × n₂
Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.
This can be shown based on the fact that the pressure of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.
