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ELEN [110]
3 years ago
13

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

is region, it will be..??deflected toward bottom of the screendeflected out of the plane of screendeflected toward top of the screendeflected into plane of screenunaffected in its direction of motion
Physics
1 answer:
muminat3 years ago
8 0

Answer:

deflected toward bottom of the screen

Explanation:

When entering the region with magnetic field, a magnetic force is exerted on the proton. This force is perpendicular to both the direction of the magnetic field and the direction of the velocity of the proton.

The direction of the force can be determined by using the right-hand rule. We have:

- Index finger: direction of the velocity of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: direction of the magnetic force --> toward bottom of the screen

So, the correct answer is

deflected toward bottom of the screen

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When an object is moved from rest at point a to rest at point b in a gravitational field, the net work done by the field depends
MaRussiya [10]
It depends on where it began<span />
5 0
3 years ago
Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the
Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

  • Final position: Mr. Llama's house;
  • Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0
3 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

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
If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.
Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so \frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{P_1}{P_2}=\frac{T_1}{T_2}

As T_1 is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0
3 years ago
A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f
svlad2 [7]

Answer:

W = 21409.2 J

Explanation:

Given,

The mass of the woman, m = 77 Kg

The length of the water slide, S = 42.6 m

The inclination of the water slide, ∅ = 42.3

The constant velocity of the women sliding, 20.3 m/s

The kinetic friction of the sliding force is given by the formula

                               Fₓ = μₓ η

Where,

                    μₓ - coefficient of kinetic friction

                     η - normal force acting on the body

Since the water slide is inclined at an angle and the person is sliding with constant velocity. The coefficient of friction becomes,

                     μₓ = tan∅

And,                η = mg cos∅

Therefore, the kinetic friction force becomes

                          Fₓ =  tan∅  mg cos∅

Substituting the given values in the above equation

                           Fₓ = 0.9 x 77 x 9.8 x 0,74

                               = 502.56 N

The work done by the kinetic friction on the person

                            W = Fₓ · S    

                                = 502.56 N x 42.6 m

                                = 21409.2 J

Hence, the work done by the friction on the woman is, W = 21409.2 J

8 0
3 years ago
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