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aleksklad [387]
3 years ago
12

How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

0.50c
B) 0.65c
C) 0.78c
D) 0.87c
Physics
1 answer:
Elan Coil [88]3 years ago
4 0

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c

Option (D) is correct.

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To balance a chemical equation we use______. <br> -coefficients <br> -subscripts
Vitek1552 [10]

Answer:

Coefficients

Explanation:

  • Coefficients are used to balance chemical equations.
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  • The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
  • <em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>
3 0
3 years ago
Many planets in our solar system (Saturn, Uranus, Neptune) have rings. If these rings are made of chunks of debris, why do they
Liono4ka [1.6K]
Generally, rings form from moons, asteroids, or comets that have disintegrated due to a collision or because they got too close to their planet (Roche Limit). ... Most of the debris, however, will not have enough energy to overcome the planet's gravity and will remain in orbit around the planet.
5 0
3 years ago
Hello,Plz Help Which conditions are necessary for rain to form?
dimulka [17.4K]
Vas happenin!
The third one makes no since because the clouds carry the rain. It isn’t always cold when it’s going to rain

The fourth one is a good one

The second one again it’s not always cold when it’s raining

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Hmmm I would go with the last one

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5 0
3 years ago
A package of aluminum foil contains 50. Ft2 of foil, which weighs approximately 7.0 oz . Aluminum has a density of 2.70 g/cm3. W
kumpel [21]

We know density = Mass / Volume

So Volume = Mass/Density

Volume =  Area * Thickness

Area = 50 ft^2 = 4.645152 m^2\\ \\Mass = 7 oz = 0.198447 kg\\ \\Density = 2.70 g/cm^3 = 2700 kg/m^3\\ \\Substituting \\ \\4.645152 *t = \frac{0.198447}{2700} \\ \\t = 1.58*10^{-5}m = 1.58*10^{-2}mm

So the approximate thickness of the foil in millimeters = 1.58*10^{-2}mm

6 0
3 years ago
A ray of light in air is incident on the mid-point of a heavy flint glass prism surface at an angle of 20º with the normal. for
Nata [24]
Assuming that you have a triangular prism, the ray of light will undergo refraction twice. The first time is the transition from air to flint glass on the entry face, and the second time is the transition from the flint glass to air from the exit face. With the available data, there are two possible solution since saying "20Âş from the normal" isn't enough information. Depending upon which side of the normal that 20 degrees is, the interior triangle will have the angles of 35, 90-r, and 55+r, or 35, 90+r, 55-r degrees where r is the angle from the normal after the 1st refraction. I will provide both possible solutions and you'll need to actually select the correct one based upon the actual geometry which I don't know because you didn't provide the figure or diagram that you were provided with.    

The equation for refraction is:  

(sin a1)/(sin a2) = n1/n2  

where  

a1,a2 = angles from the normal to the surface.  

n1,n2 = index of refraction for the transmission mediums.    

For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So  

(sin a1)/(sin a2) = n1/n2  

(sin 20)/(sin a2) = 1.00029/1.60  

0.342020143/(sin a2) = 0.62518125  

0.342020143 = 0.62518125(sin a2)  

0.547073578 = sin a2  

asin(0.547073578) = a2  

33.16647891 = a2    

So the angle from the normal INSIDE the prism is 33.2Âş. The resulting angle from the surface of the entry face will be either 90-33.2 or 90+33.2 depending upon the geometry. So the 2 possible triangles will be either 35Âş, 56.8Âş, 88.2Âş or 35Âş, 123.2Âş, 21.8Âş. with a resulting angle from the normal of either 1.8Âş or 68.2Âş. I can't tell you which one is correct since you didn't tell me which side of the normal the incoming ray came from. So let's calculate both possible exits.    

1.8Âş  

(sin a1)/(sin a2) = n1/n2  

(sin 1.8)/(sin a2) = 1.6/1.00029  

0.031410759/(sin a2) = 1.599536135  

0.031410759= 1.599536135(sin a2)  

0.019637418= sin(a2)  

asin(0.019637418) = a2  

1.125213477 = a2    

68.2Âş  

(sin a1)/(sin a2) = n1/n2  

(sin 68.2)/(sin a2) = 1.6/1.00029  

0.928485827/(sin a2) = 1.599536135  

0.928485827 = 1.599536135(sin a2)  

0.58047193 = sin a2 

 asin(0.58047193) = a2 

 35.48374252 = a2   

 So if the interior triangle is acute, the answer is 1.13Âş and if the interior triangle is obtuse, the answer is 35.48Âş
5 0
3 years ago
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