Answer:
Explanation:
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .
As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.
If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.
Answer:
a 
10.6 m/s²
Explanation:
Since F = ma (Force = mass * acceleration), acceleration would be...
a = F/m
a = 302 N/28.6
a 10.6 m/s²
It is the red line the purple line carries the most the green line is the largest and the smallest is sky blue
Answer:
D. TA < TB
Explanation:
From general gas equation, we know that:
PV = nRT
PV/R = nT
where,
P = pressure of gas
V = volume of gas
R = General gas constant
T = temperature of gas
n = no. of moles of gas
<u>FOR CYLINDER A</u>:
PV/R = (nA)(TA) _____ eqn (1)
<u>FOR CYLINDER B</u>:
PV/R = (nB)(TB) _____ eqn (2)
Because, Pressure, Volume are constant for both cylinders.
Comparing eqn (1) and (2)
(nA)(TA) = (nB)(TB)
It is given that the amount of gas in cylinder A is twice as much as the gas in cylinder B. This means the number moles in cylinder A are twice as much as no. of moles in cylinder B.
nA = 2(nB)
using this in eqn:
2(nB)(TA) = (nB)(TB)
TA = (1/2)(TB)
<u>TA = 0.5 TB</u>
Therefore it is clear that the correct option is:
<u>D. TA<TB</u>
ANSWER
My answer is in the photo above
