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vlada-n [284]
3 years ago
9

The Starship Enterprise speeds past an asteroid at 0.920c. If an observer on the asteroid sees 10.0 seconds pass on her watch, h

ow long would that time interval be if measured by an observer on the Enterprise?
Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

25.52 seconds

Explanation:

Speed of space ship = 0.92 c = v

c = Speed of light

Time observed from asteroid = Δt = 10 seconds

Time dilation

\Delta t'=\frac{\Delta t}{\sqrt{1-\frac {v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-\frac {0.92^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-0.92^2}}\\\Rightarrow \Delta t'=25.52\ s

∴ Time interval be if measured by an observer on the Enterprise would be 25.52 seconds

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Se puede apelar la circunstancias para justificar una decisión que afecta a otras personas
VladimirAG [237]

Answer:

No

Explanation:

tratar de justificar nuestras propias decisiones es algo totalmente innecesario, ya que todos tenemos derecho a tomar decisiones que en ocasiones pueden afectar indirectamente a otras personas y no hay nada de malo en ello.

Por otro lado, si está tomando una acción o tomando una decisión únicamente con la intención de afectar a otras personas, su acción no tiene justificación alguna.

8 0
3 years ago
What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria
Likurg_2 [28]

Answer:

Approximately \rm 11.2 \; N \cdot kg^{-1} at that distance from the center of the planet.

Option A) The low value of g near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of g on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

\displaystyle \frac{G \cdot M \cdot m}{R^2},

where

  • G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2.
  • M is the mass of the planet, and
  • m is the mass of the object.

To find an equation for g, divide the equation for gravity by the mass of the object:

\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}.

In this case,

  • M = 5.68\times 10^{26}\; \rm kg, and
  • R = 5.82 \times 10^7\; \rm m.

Calculate g based on these values:

\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}.

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for g:

\displaystyle g = \frac{G \cdot M}{R^2}.

The mass of the planet is in the numerator. If two planets are of the same size, g would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since R is in the denominator of g, increasing the value of R while keeping M constant would reduce the value of g. That explains why the value of g near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately 9.81\; \rm N \cdot kg^{-1}.

As a side note, 5.82\times 10^7\rm \; m likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of g near the cloud tops of Saturn is approximately \rm 11.2 \; N \cdot kg^{-1}.

6 0
3 years ago
The air pressure inside the tube of a car tire is 430 kPa at a temperature of 13.0 °C. What is the pressure of the air, if the t
KengaRu [80]

Answer:

The pressure of the air is 499.91 kPa.

Explanation:

Given that,

Initial pressure = 430 kPa

Temperature = 13.0+273=286 K

Final temperature = 59.5+273=332.5 K

We need to calculate the final pressure

Using relation of pressure and temperature

At constant volume,

\dfrac{P'}{P}=\dfrac{T'}{T}

\dfrac{P'}{430}=\dfrac{332.5}{286}

P'=\dfrac{332.5}{286}\times430

P'=499.91\ kPa

Hence,The pressure of the air is 499.91 kPa.

7 0
3 years ago
A trolley has a mass of 1.2 kg and a speed of 4.5 m/s. The trolley crashes into a stationary trolley of mass 0.8 kg. On impact t
attashe74 [19]

(a) The momentum of the first trolley is 5.4 kgm/s

(b) The velocity of the trolleys after impact is 2.7m/s

<u>Explanation:</u>

Given:

Mass, m₁ = 1.2kg

Velocity, v₁ = 4.5m/s

Mass, m₂ = 0.8kg

v₂ = 0

(a) Momentum of the trolley before impact, p

We know:

Momentum = mass X velocity

p = 1.2 X 4.5

p = 5.4 kgm/s

Therefore, the momentum of the first trolley is 5.4 kgm/s

(b) Speed of the trolleys after impact, v = ?

During collision, the momentum is conserved.

So,

m₁v₁ + m₂v₂ = (m₁ + m₂)v

(1.2 X 4.5) + (0.8 X 0) = (1.2 +0.8) X v

5.4 + 0 = 2v

v = 2.7m/s

Therefore, the velocity of the trolleys after impact is 2.7m/s

8 0
3 years ago
Why doesn’t change in a substance during a physical change?
iris [78.8K]

Answer:

there is no change because a physical change only changes the physical appearance of a substance

Explanation:

a chemical change is the only change that can change a substance since it changes the chemical properties of the substance, hence making a new substance

3 0
3 years ago
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