Answer:
did you have the same answer to get the best
Answer:
7.16 m /s
Explanation:
The depth of the small pipe attached with the side wall of tank from the surface of water
h =( 3.1 - .48 )m
= 2.62 m
velocity of flow of water= √ 2 g h
= √ 2 x 9.8 x 2.62
= 7.16 m /s
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The density is 4.76 gcm^-3
and if mass is in kg then density is equal to 4.76*10^-3
Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
Kinetic energy = 1/2 m v²
If we reduce the mass by half > m/2
Kinetic energy = 1/2 m/2 v²
We should know that 1/2 × 1/2 = 1/4
So kinetic energy will be :
1/4 × m × v²
