Physics High School

How do I calculate how many meters are in 7.2 light years?
The exact question is:
Calculate in meters the distance between a galaxy and the Earth if the distance is equal to 7.2 light years.

Answers

Answer 1

Answer:

68.2 Quadrillion meters

Explanation:

A lightyear is the distance that light travels in one year.

Speed of light is [tex]3*10^8\ m/s[/tex]

So light covers 300,000,000 meters in one second.

One year has 31536000 seconds so , light covers

[tex]9.461*10^{15}\ meters\ in\ one\ year[/tex]

so 7.2 light years is

[tex]7.2*(9.461*10^{15})\\6.82*10^{16}[/tex]

so 7.2 light years is

6.82 x 10^(16) meters or

68.2 Quadrillion meters

Related Questions

Help me pretty please with a cherry on top!!

Answers

it doesnt show anything for me..


explaination :: cornje

Answer:

A. im not for sure hope it helps

Explanation:

Help please!!!
if superman at 90kg jumps a 40m building in a single bound how much work does superman perform

Answers

Answer:

5 years worth of work (aka all of the homework i currently have)

what power is transmitted by 2A flowing across 5V​

Answers

Answer:

[tex]\huge\boxed{10\:watts}[/tex]

Explanation:

Power = Current × Voltage

P = I × V

P = 2 × 5

P = 10 watts

Electric power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.

Answers

Answer:

[tex]\eta_{turbine} = 0.777 = 77.7\%[/tex]

[tex]\eta_{combined} = 0.728 = 72.8\%[/tex]

Explanation:

First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:

Input Power = [tex]P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW[/tex]

Now, for turbine efficiency:

[tex]\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%[/tex]

for generator efficiency:

[tex]\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%[/tex]

Now, for combined efficiency:

[tex]\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%[/tex]

An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?

Answers

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = [tex]\frac{12}{49}[/tex]   = 0.25m/s²

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