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PIT_PIT [208]
3 years ago
7

How many minutes would it take a light wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28

million miles).
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

t=2.51min

Explanation:

The time taken by the light to travel a given distance is defined as:

t=\frac{d}{c}

Here c is obviously the speed of light. Now we convert the average distance form Venus to Earth to meters:

28*10^{6}mi*\frac{1609.34}{1mi}=4.51*10^{10}m

Finally, we calculate the minutes taken by the light to travel from Venus to Earth:

t=\frac{4.51*10^{10}m}{3*10^8\frac{m}{s}}\\t=150.33s*\frac{1min}{60s}=2.51min

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The white dwarf that remains when our sun dies will be mostly made of
expeople1 [14]

Answer:

Carbon and oxygen

Explanation:

White dwarfs are the stars which have used all their hydrogen and helium fuel and now exists with only carbon and oxygen in their core. Their size reduces up to one hundredth times of the size of their sun in early stages and yet they possess the same mass.  

Due to loss of fuels and impact of gravity, a young star collapses on itself leading to formation of dwarf star.

 

7 0
3 years ago
Light can be transmitted through long fiber optic strands because of
wariber [46]
Total internal reflection. Although I'm not 100%


3 0
3 years ago
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A sky diver, mass 90 kg, reaches a terminal velocity of 60 m/s. What is the approximate magnitude of the force of air resistance
Komok [63]

Answer:

The approximate magnitude of the force of air resistance is 540 N.

Explanation:

6 0
2 years ago
How fast must a 350 gram softball be traveling in order to have a momentum of .600 kg
BartSMP [9]
Momentum is mass times velocity. So here we can just substitute in our givens and solve for velocity.

.600kg*m/s/.350kg=1.71m/s

Hope this helps! Thank you!
4 0
4 years ago
A student used apparatus as shown above. The beaker held 750 g of a liquid. The current from the power supply was 1.8 A and the
liq [111]

Answer:

2365.71\ \text{J/kg}^{\circ}\text{C}

Explanation:

V = Voltage = 230 V

I = Current = 1.8 A

\Delta T = Temperature change = (40-12)^{\circ}\text{C}

t = Time taken = 2 minutes

m = Mass of liquid = 750 g

c = Specific heat capacity of the liquid

Energy required to heat the water is equal to the heat released due to current passing

mc\Delta T=VIt\\\Rightarrow c=\dfrac{VIt}{m\Delta T}\\\Rightarrow c=\dfrac{230\times 1.8\times 2\times 60}{0.75\times (40-12)}\\\Rightarrow c=2365.71\ \text{J/kg}^{\circ}\text{C}

The specific heat capacity of the liquid is 2365.71\ \text{J/kg}^{\circ}\text{C}

6 0
3 years ago
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