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sergeinik [125]
4 years ago
15

infer whether a planet with active volcanoes would have more craters than a planet without active volcanoes

Physics
1 answer:
just olya [345]4 years ago
7 0
I predict that the planet with active volcanoes will have more craters, due to the volcanic explosions, as opposed to a planet without active volcanoes.
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How does a water molecule transfer from the ocean to form part of a cloud in the atmosphere?
Luden [163]
D.
The molecule escapes the ocean via evaporation and then becomes part of a cloud via condensation.
8 0
4 years ago
2.) The speed of light in a vacuum is c = 300,000 km / s. The light of Sol takes 8 minutes and 19 seconds to reach Earth. Calcul
Vinvika [58]

Answer:

149,700,000 km   (= 1.50 x 10⁸ km)

Explanation:

Given,

Speed of light, c = 300,000 km/s

Time Taken = 8 min 19 s = 499 seconds

Recall, Distance = Speed x Time

= 300,000 km/s   x 499 s

= 149,700,000 km

= 1.50 x 10⁸ km

3 0
3 years ago
Read 2 more answers
The half-lives for the radioactive decay of various elements are listed below.
myrzilka [38]

Answer:

D ko alam pasensya ka na ha

4 0
3 years ago
A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0
3 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
3 years ago
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