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lana [24]
3 years ago
13

At the Equator near Earth’s surface, the magnetic field is approximately 82.2 µT northward and the electric field is about 143 N

/C downward in fair weather. A electron travels with an instantaneous velocity of 2.42 × 106 m/s directed to the east in this environment. The acceleration of gravity is 9.8 m/s 2 . Find the magnitude of the gravitational force on the electron. Answer in units of N.
Physics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

Magnitude of gravitational force of the electron= 3.74×10^12N

Explanation:

Felectron= Fgravitational

Therefore:

Felectron/Fgravitational = kq^2/r^2 ×(r^2/Gm^2)

= kq^2/Gm^2

Where G= gravitational constant

m = mass of electron=9.1×10^-31

K= 8.9x10^9

q= 1.6×10^-19

Magnitude of gravitational force= (8.9×10^9)×(1.6×10^-19)^2/(6.7×10^-11)×(9.1×10^-31)

=( 2.2784×10^-28) / ( 6.097×10^-41)

= 3.74×10^12N

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60 POINTS ANSWER CORRECTLY
ohaa [14]

Answer:

C ) 1.53

Explanation:

The critical angle of a material is given by the formula

sin c = \frac{1}{n}

where

c is the critical angle

n is the refractive index

This formula is valid if the second medium is air (which is the case of the problem).

In this problem, we know the critical angle:

c=40.8^{\circ}

Therefore we can rearrange the equation to find the refractive index:

n=\frac{1}{sin c}=\frac{1}{sin 40.8^{\circ}}=1.53

5 0
4 years ago
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Marrrta [24]
Loooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool
7 0
3 years ago
Can you make the work output of a machine greater than the work input?
ExtremeBDS [4]

Answer:

It is<em> impossible</em> to construct a machine which produces the <em>work output greater than the work input.</em>

Let us consider the II law of thermodynamics.

According to Kelvin Plank's statement any engine/machine does not give hundred percent efficiency. And violating the PMM-II(Perpetual motion of machine II kind), Always some amount of energy transferred  to the sink or surroundings.

Therefore

                     W(ouput) = Q₁-Q₂

There are many reasons to lower the work output, just for an example friction between the mating parts reduces the work output.

5 0
3 years ago
Read 2 more answers
Need help in the middle one
kondor19780726 [428]

Answer:

Guysi hate math answer this guy plsss ssss

7 0
3 years ago
Can someone explain to me #4.
nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
4 0
3 years ago
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