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ss7ja [257]
3 years ago
15

In which of the following situations does gravity supply a centripetal force?

Physics
1 answer:
kkurt [141]3 years ago
7 0
Space shuttle orbiting the earth 
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A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m
Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

C_{1} = \epsilon_{0}*\frac{A}{d_{1}}

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m    

A: is the plate area = 865 mm² =  8.65x10⁻⁴ m²

C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F      

Similarly, C₂ is:

C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F

Now, V₂ can be calculated by finding the initial charge (q₁):

q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C

Since, q₁ is equal to q₂, V₂ is:

V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V

Finally, we can find the work:

W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0
3 years ago
Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.
iren [92.7K]

Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

For F_n

F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

8 0
3 years ago
peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie pla
zavuch27 [327]

Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let \theta is the angular distance that the pie plate has moved through.

It is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{183}{4.5}  

\theta=40.66\ radian

Since, 1 radian = 57.29 degrees

\theta=2329.64\ degrees

Since, 1 radian = 0.159155 revolution

\theta=6.47124\ revolution

Hence, this is the required solution.

5 0
3 years ago
A 10.0 kilogram mass of iron on earth exerts a downward force of 98 Newtons. On the moon this same 10.0 kilograms of iron is wei
Len [333]

Answer: D) The gravity on the moon is one sixth the gravity of Earth.  

Explanation:

6 0
3 years ago
Read 2 more answers
Helpp
frosja888 [35]
B, refraction I believe.
5 0
2 years ago
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