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julsineya [31]
3 years ago
8

Jake is bowling with an 8 kg bowling ball roll the ball at 7 Ms and it hits 1 stationary pain with a mass of 2kg the pain goes f

lying forward at 12 Ms how fast is the ball now traveling
Physics
1 answer:
bearhunter [10]3 years ago
8 0

Solution

In this question we have given

mass of bowling ball=8kg

Velocity of bowling ball= 7m/s

mass of pain=2kg

velocity of pain=12m/s

In this case, law of conservation of momentum will be applied. According to law of conservation of momentum,

initial momentum of system= final momentum of the system

Total initial momentum of the system =mass of bowling ball* Velocity of bowling ball

=8 kg x 7 m/s

= 56 kg m/s.

momentum of the pain after getting hit is =mass of pain* velocity of pain

=2kg*12m/s

=24 kg m/s.

So the momentum of the bowling ball after hitting the pain= 56 kg m/s - 24 kg m/s

= 32 kg m/s

velocity of the bowling ball after hitting the pain = momentum of the bowling ball after hitting the pain/mass of bowling ball

=32 kg m/s / 8 kg

= 4 m/s

Therfore, velocity of the bowling ball after hitting the pain is 4m/s


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Answer:

Option (3)

Explanation:

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He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.

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6 0
3 years ago
A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
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Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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In kw it's going to be 600/1000=0.6kw

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