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
Answer:
answer is 1000 N
formula used-
<em><u>F= m x (v-u/t)</u></em>
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<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>
By applying Newton's second law of motion;
ma = mg - T
Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.
For the current case, the velocity is constant therefore,
a = 0
Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N
Tension in the cable is 1128.15 N.