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Bond [772]
3 years ago
6

A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with

a velocity of 4.0 m/s [W]. Ignore the substantial friction real objects in water experience.
A) What is the momentum of the girl as she is diving?
B) What is the momentum of raft as the girl is diving?
C) What is the final speed of the raft just after the girl dives?
Physics
1 answer:
lord [1]3 years ago
6 0
Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s

B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]
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What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

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radius r=0.20 m

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(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

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3 0
3 years ago
Please help on this one?
djyliett [7]
It’s C
Cause Impulse is found by multiplying the force and change in time (which is simply time)
So if you rearrange the equation for time you end up dividing Impulse by force.
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