Answer:
10
Explanation:
10 is the answer because you divide 50 and 5 and your answer is 10.
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Answer:
The force is applicable according to the newton's third law of motion but the force on the engine is compensated in the form of stess on the fixed parts and rigid links whereas the wheel is free to roll.
Explanation:
This interpretation that the engine applies a force on the train and the train also applies an equal force on the engine and hence the train should not move is wrong because the engine imparts a rotational force in the form of torque on the train and the train imparts an equal force on the engine in the opposite direction but the engine is fixed like a structure on the chassis of the train and consists of rigid links which resist the motion and deformation as compared to the relative motion between the wheels and the rail tracks.
<span>The weight of the spacecraft keeps changing.
</span>
<span>The mass of the spacecraft remains the same.
These are the correct answers</span>
<span>at maximum height the final velocity will be 0
using v=u+at and resolving vertically we get
v=0.6+(-9.81)t
v=0.6-9.81t
0=0.6-9.81t
9.81t=0.6
t=0.6/9.81
t=0.061 to 3sf
Now we need to resolve horizontally to find the horizontal distance
using s=ut+1/2at^2
However we now need the total time taken for the projectile travel and return to the ground. We can assume the time taken for the projectile to reach its maximum height and return to the ground is the same therefore
the total time is 2 x 0.061=0.122seconds. They'll be now horizontal acceleration in this case scenario therefore
Hence s=ut+1/2at^2
since a=0
s=ut
s=0.6 x 0.122
s=0.073m
</span>
Answer:
Velocity of skater after throwing the snowball is 2.57 m/s
Explanation:
Given :
Mass of skater, M = 62.2 kg
Mass of snowball, m = 0.145 kg
Velocity of snowball relative to ground, v = 39.3 m/s
Consider v₁ be the velocity of skater after throwing the snowball.
According to the problem, initially the velocity of skater and snowball is same. So,
Velocity of skater before throwing snowball, u = 2.66 m/s
Applying conservation of momentum,
Momentum before throwing snowball = Momentum after throwing snowball
(M + m) u = Mv₁ + mv
Substitute the suitable values in the above equation.
v₁ = 2.57 m/s