Answer:
O ksm/s
Explanation:
before collision,
Velocity =0
So,momentum of the bottle before collision=mass ×velocity
=mass×0
=0 kgm/s
Answer:
a) 200A
b) 10.2V
c) 2.04kW
d)
I=80A
V=4.08V
P=0.326kW
Explanation:
Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

Where R is the equivalent resistance of the resistors in series
![R=0.0510+0.0090=0.0600[ohm]](https://tex.z-dn.net/?f=R%3D0.0510%2B0.0090%3D0.0600%5Bohm%5D)

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

The power dissipated supplied to the motor is given by:

now solving adding a 0.0900 ohm resistor:



Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)

For the bottom of window (position B)


We also have

Solving

So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s
A wave will "break" because the bottom interferes with its oscillatory motion. Breaking of waves may occur anywhere that the amplitude is sufficient, including in mid-ocean. When waves enter shallow water they break because the motion of water in lower part of the wave nearest the bottom is slowed by friction so that their oscillation is faster than its supporting portion at the bottom. Thus, the wave collapses forward and breaks.