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forsale [732]
2 years ago
9

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

g at close to the speed of light. Suppose you wish to visit the red giant star Betelgeuse, which is 430 ly away, and that you want your 20,000 kg rocket to move so fast that you age only 38 years during the round trip.
a)How fast must the rocket travel relative to earth?

B)How much energy is needed to accelerate the rocket to this speed?

c)The total energy used in the United States in the year 2005 was roughly 1.0 *10^20 J. Compare the rocket's energy to this value by computing the ratio Energy of the rocket/ Energy used by the US.
Physics
1 answer:
Vika [28.1K]2 years ago
7 0

Answer:

Explanation:

Let the required velocity of rocket be v .

We shall use the formula of time dilation to find the velocity of rocket .

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2} } }

t = 430

t' = 38

430=\frac{38}{\sqrt{1-\frac{v^2}{c^2} } }

\sqrt{1-\frac{v^2}{c^2} } }=\frac{38}{430}

1-\frac{v^2}{c^2} = .0078

\frac{v^2}{c^2} =.9922

\frac{v}{c} = .996

v  = .996 x 3 x 10⁸ m /s

= 2.988 x 10⁸ m /s

B )

Kinetic energy of rocket

= 1/2 m v²

= .5 x 20000 x (2.988 x 10⁸ )²

= 8.9 x 10²⁰ J .

C )

This energy is 8.9 times the energy requirement of United states in the year 2005 .

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8m/s

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a=32/4

a=8 m/s

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
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Answer:

\text{heat loss} = 24864.05 \  W/m^2

Explanation:

If

  • T_1, T_2 are temperatures of gasses and liquid in Kelvins,
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  • h_1, h_2 are convection coefficients gas and liquid in W/m^2 \cdot K,
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  • and k_1, k_2 are thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

using known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Using the rate equation :

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Similarly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature distribution is shown in the attached image

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Answer:

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A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

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False, the acceleration is constant in magnitude throughout the motion

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True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

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