Answer:
Speed of the helium after collision = 246 m/s
Explanation:
Given that
Mass of helium ,m₁ = 4 u
u₁=598 m/s
Mass of oxygen ,m₂ = 32 u
u₂ = 401 m/s
v₂ =445 m/s
Given that initially both are moving in the same direction and lets take they are moving in the right direction.
Speed of the helium after collision = v₁
There is no any external force on the masses that is why the linear momentum will be conserve.
Initial linear momentum = Final linear momentum
P = m v
m₁u₁+m₂u₂ = m₁v₁+m₂v₂
598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445
v₁ = 246 m/s
Speed of the helium after collision = 246 m/s
I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.
This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)
The answer is momentum.
Momentum is (mass x speed). Change in momentum is (force x time).
No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.
Explanation:
The given value of P is as follows.
P = 1.06F + 22.18, 
or, P' = 1.06
As p' is defined and non-zero. Hence, only critical points are boundary points.
For F = 10, the value of P will be calculated as follows.
P = 
= 32.78
For F = 70, the value of P will be calculated as follows.
P = 
= 96.38
Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.
Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.
C. Figure 11 and figure 1v are both compounds