Answer: C.
Explanation:
For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.
Now, the stored energy can be written as:
E = (1/2)*Q^2/C
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.
![N_x=86.62N](https://tex.z-dn.net/?f=N_x%3D86.62N)
- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.
![N_y=F_V=mg-Tsin59\\](https://tex.z-dn.net/?f=N_y%3DF_V%3Dmg-Tsin59%5C%5C)
- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.
![F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N](https://tex.z-dn.net/?f=F_H%3DN_x%3DTcos59%5C%5C%5C%5CT%3D%5Cfrac%7BF_H%7D%7Bcos59%7D%20%3D%5Cfrac%7B86.62%7D%7B0.51%7D%3D%20169.84N)
- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,
![N_y= (40*9.8)-(169.8*sin59)=246.4N](https://tex.z-dn.net/?f=N_y%3D%20%2840%2A9.8%29-%28169.8%2Asin59%29%3D246.4N)
- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,
![N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N](https://tex.z-dn.net/?f=N%3D%5Csqrt%7BN_x%5E2%2BN_y%5E2%7D%20%3D%5Csqrt%7B%2886.62%29%5E2%2B%28246.4%29%5E2%7D%3D261.12N)
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
Learn more about the tension here:
brainly.com/question/28106871
#SPJ1
I would say that insofar as the two stars temperatures are presumably closely related to their luminosity, that the blue star at 156,100 k compared to 3000k for the red star then the blue star has a luminosity of 52 times that of the red star.
<span>Answer:
sin(incidence)/sin(refraction) = n_refraction/n_incidence
sin(50) / sin(x) = 1.5 / 1
sin(50)/1.5 = sin(x)
sin(x) = 0.511
x = 30.71o
B]
50 degrees, same as the angle going in.
You can show that by reversing the steps in A.
sin(30.7)/sin(x) = 1/1.5
C]
The glass is 5 cm thick.
The reference angle = 30.7o
Tan(30.7) = displacement / thickness
Tan(30.7) = x / 5
5*sin(30.7) = x
x = 2.97 cm which is the displacement.</span>
14.59390 kg hope it helps