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noname [10]
3 years ago
15

a 60g tennis ball travelling at 30m/s hi a wall and bounce back at 20m/s.calculate (a)momentum of the ball before impact (b)mome

ntum of the ball after impact​
Physics
1 answer:
frez [133]3 years ago
6 0

Explanation:

the formula for momentum is denoted by p=mv where p is momentum, m is mass and v is velocity. thus, the velocity before impact would be 0.060 x 30 = 1.8 kg/ms

the second one would just be 0.060 x 20 0.72kg/ms

I'm not 100 percent sure this is correct but yeah

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3 years ago
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
3 years ago
Which of the following is an example of a noninfectious disease?
Scorpion4ik [409]
D lung cancer is not infectious
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3 years ago
Read 2 more answers
A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

6 0
3 years ago
Metals can be described as?
luda_lava [24]

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