Answer:
1 and 2
Explanation:
It is given that, force exerted by air is negligible in any way.
Also, it is given that channel is in the shape of a segment of a circle with its center at O.
When it is within the frictionless channel at position 'Q',
A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.
However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.
The recoil velocity of cannon is (4) 5.0 m/s
Explanation:
We can find the recoil velocity from the law of conservation of momentum.
The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.
Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec
m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.
So recoil velocity of cannon v2 is given by,
v2 = -(m1÷m2)v1
v2 = -(100÷500)15
v2 = -5 m/s
where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.
Hence, option (4)5.0 m/s is the correct answer.
Answer:
Answer: C
explanation:
They could be same or different
ie:
|-5|,|-5| = 5,5
|-5|,|5| = 5,5
Answer:
g = 0.98 m
Explanation:
given data:
upper portion weight 438 N and its center of gravity is 1.17 m
upper legs weight 144 N and its center of gravity is 0.835 m
lower leg weight 88 N and its center of gravity is 0.270 m
position of center of gravity of whole body can be determine by using following relation
where 
is mass of respective part and 
is center of gravity
g = 0.98 m
The centripetal force and centripetal acceleration
both point toward the center of the circular path.
