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Katyanochek1 [597]
2 years ago
9

Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B

Physics
1 answer:
pogonyaev2 years ago
8 0

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
Because P and S waves travel faster through the Earth's mantle than through Earth's crust, scientists know that the mantle is __
larisa [96]
B. liquid and less denser
6 0
2 years ago
Read 2 more answers
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
2 years ago
The noise level coming from a pig pen with
sweet [91]

Answer:

The decibel of the remaining pigs is 51.5 dB.

Explanation:

Decibel (dB) is a unit of measure of the intensity of a given sound.

Number of pigs = 199, noise level = 74.3 dB.

Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:

                       I    \alpha  N

                       I = kN

where k is the constant of proportionality.

⇒                    k = \frac{I}{N}

                         = \frac{74.3}{199}

                      k = 0.3734

When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.

Thus, the becibel level (I) of the remaining pigs can be determined by:

                  I = kN

                    = 0.3734 × 138

                   = 51.53 dB

The becibel level (I) of the remaining pigs is 51.53 dB.

6 0
3 years ago
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe
stiks02 [169]

Answer:

6.0 × 10^{11} W/m^{2}

Explanation:

From Wien's displacement formula;

Q = e AT^{4}

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = \frac{Q}{A} = eT^{4}

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × (1000)^{4}

                             = 0.6 × 1.0 × 10^{12}

                             = 6.0 × 10^{11} \frac{W}{m^{2} }

Therefore, the emissive intensity coming out of the surface is 6.0 × 10^{11} W/m^{2}.

3 0
2 years ago
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