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blondinia [14]
3 years ago
9

A of mass 5.0 kilograms moving at 20. meters per second collides with ball B of unknown mass moving at 10. meters per second in

the same direction. After the collision, ball A moves at 10. meters per second and ball B at 15 meters per second, both still in the same direction. What is the mass of ball B
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

Mass of B is 10 kg

Explanation:

M1v1a + m2v1b = m1v2a + m2v2b

Let m2 = x

5*20 + x10 = 5*10 + x15

100 + x10 = 50 + x15

50 = x5

X= 10

Mass of B is 10 kg

M1 = mass of A

V1a = initial velocity of A

V2a = final velocity of A

M2 = mass of B

V1b = initial velocity of B

V2b = final velocity of B

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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
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Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
Over the last several decades scientists have considered the problem of nonrenewable natural resources, such as fossil fuels. Hu
sesenic [268]

The statement "Good locations for turbines are limited" describes a drawback to wind energy.

Answer: Option B

<u>Explanation:</u>

Wind energy is one of the most useful and efficient renewable energy sources. But nothing is ideal in this universe and the same thing applies for wind energy also. The generation of electricity from wind energy requires setting up of turbines.

And these turbines can be set up in plane areas which is free from any disturbance except wind flow. In open area and flat plane surface only the turbines can rotate freely with the effect of wind.

But regions where the wind flow is minimum due to snow formation like the northern region of earth, the turbines cannot be set up there. So the locations for setting up of turbines are limited for good outcome in wind energy. This is one of the drawback of wind energy.

8 0
3 years ago
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0
3 years ago
Select the false statement. A. Only materials that have never been part of a living thing can be recycled in nature. B. Matter f
astra-53 [7]

Option A is the false statement.

Only materials that have never been part of a living thing can be recycled in nature.

In actual all living and non living thing can be recycled in nature.

3 0
3 years ago
An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen
Ad libitum [116K]

Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.

Explanation:

The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:

F = BqVSinØ

If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field

According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.

6 0
3 years ago
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