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enot [183]
3 years ago
11

Which of the following is not a subatomic particle?

Physics
1 answer:
Fynjy0 [20]3 years ago
7 0
D. Nucleus because it is not a part of the group.
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A 250 - ω resistor is connected in series with a 4.80 - μf capacitor. the voltage across the capacitor is vc=(7.60v)⋅sin[(120rad
galben [10]
<span>553 ohms The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is Xc = 1/(2*pi*f*C) where Xc = Reactance in ohms pi = 3.1415926535..... f = frequency in hertz. C = capacitance in farads. I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance. The 120 rad/s frequency is better known as 60 Hz. Substitute known values into the formula. Xc = 1/(2*pi* 60 * 0.00000480) Xc = 1/0.001809557 Xc = 552.6213302 Rounding to 3 significant figures gives 553 ohms.</span>
6 0
3 years ago
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Mila both 2 kg of cheese in the market to be used for her Mac and cheese recipe. How many pounds does a 2.00 kg of cheese weigh?
gogolik [260]

Answer:

The answer is 4 pounds

Explanation:

The explanation is that 1 kilogram is equal to 2 pounds so multiply the kilogram with the 1 pound

8 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
Which statement about the carbon cycle is most true?
scZoUnD [109]
B answr is right there
3 0
3 years ago
Eddie set up a science experiment with plants. He bought three small tomato plants, all about the same size and height, and set
Elza [17]

Answer:

The frequency of watering

Explanation:

The manipulated variable in this case is the frequency of adding water to the experimental plants.

While the first plant can be said to have a watering frequency of a day, the second had a watering frequency of 2 days while the third plant had a watering frequency of 3 days.

The experiment must have been set up to determine the effects of frequency of watering on the growth of tomato plants.

8 0
3 years ago
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