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zzz [600]
3 years ago
12

Light requires 4.5 years to travel from the nearest star to earth. If we could travel there in a spaceship going 90% of the spee

d of light, the trip would require 5.0 years according to clocks on earth. How much time would pass for the passengers in the ship
Physics
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Time according to earth clock (T0) = 0.22 years (Approx)

Explanation:

Given:

Time taken by light = 4.5 years

Time taken by ship = 5 years

Speed of light = c

Speed of ship (v) = 90% of c = 0.9c

Find:

Time according to earth clock (T0) = ?

Computation:

Time dilation is ,

T(Difference) = \frac{T0}{\sqrt{1-\frac{v^2}{c^2} } }\\\\ (5-4.5)= \frac{T0}{\sqrt{1-\frac{(0.9c)^2}{c^2} } }\\\\ 0.5=\frac{T0}{\sqrt{1-0.81} }\\\\T0 =0.2179

Time according to earth clock (T0) = 0.22 years (Approx)

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(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 108 m. Find
Goshia [24]

Answer:

It took 1.28 seconds to his voice to reach the Earth via radio waves.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

Light propagates as electromagnetic wave in vacuum with a speed of 3x10^{8}m/s. Therefore, radio waves will have in vacuum the same speed.

Then, to know the time that it took for its voice, the next equation can be used:

c = \frac{d}{t}  (1)

Where c is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

t = \frac{d}{c} (2)

t = \frac{3.85x10^{8} m}{3x10^{8}m/s}

t = 1.28s

Hence, it took 1.28 seconds to his voice to reach the Earth via radio waves.

7 0
3 years ago
What percentage does each parent contribute to a child's genotype?​
bearhunter [10]

Each parent contribute 50 percent. A person have 46 genes and 23 comes from the mother and 23 comes from the father

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3 years ago
Read 2 more answers
It is 5.5 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h
stealth61 [152]

Explanation:

Displacement = 5 km

A.

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

= 25/9 m/s

Remember,

700 watt = 700 J/s

Velocity = displacement/time

Time = 5000/(25/9)

= 1800 s

Energy = power * time

= 700 * 1800

= 1,260,000

= 1260 kJ

B.

Converting km/h to m/s,

3 km/h * 1000 m/1 km * 1 h/3600 s

= 5/6 m/s

290 watt = 290 J/s

Velocity = displacement/time

Time = 5000/(5/6)

= 6000 s

Energy = power * time

= 290 * 6000

= 1,740,000

= 1740 kJ

C.

Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.

6 0
3 years ago
What is the impulse of a 3kg object accelerating from rest to 12m/s?
ale4655 [162]
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).

The answer is 36 kg m/s
6 0
3 years ago
Percent Yield Lab Report
Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0
2 years ago
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