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Art [367]
3 years ago
8

An astronaut in a spacecraft looks out her window and observes a comet travel in the opposite direction at a relative speed of 2

37 m/s. The velocity of the spacecraft is 114 m/s directly away from the sun. What is the comet's velocity relative to the sun? Assume that motion away from the sun is in the positive direction.
Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0
The velocity of the spacecraft with respect to the Sun is
v_s = +114 m/s
the velocity of the comet in the reference system of the spacecraft is
v_c' = 237 m/s

So, the velocity of the comet in the reference system of the Sun will be
v_c = v_c'-v_s = -237 m/s - 114 m/s=-351 m/s
And the negative sign tells us that the comet is directed toward the Sun.
Umnica [9.8K]3 years ago
4 0

Answer: The relative velocity of the comet with respect to sun is '-351 m/s'.

Explanation:

Let the magnitude of the velocity of the spaceship moving away from the sun be |v_s|=114 m/s

Let the magnitude of the velocity of the comet moving towards the sun be |v_c| m/s

Assuming the direction away from the sun to be positive then :

The velocity of the space ship :v_s=+114 m/s

The  velocity of the comet :-v_c

v_r = relative velocity of spaceship with respect to comet

v_r=-237 m/s=v_s+(-v_c)=114 m/s-v_c

v_c=351 m/s

Velocity of the comet.v_c=351 m/s

v_R = Relative velocity of comet with respect to sun:

v_R=v_s+v_c=0 m/s+(-351 m/s)=-351 m/s

v_s= 0 m/s(Sun is fixed at one point)

The negative sign means that comet is moving towards the sun.The relative velocity of the comet with respect to sun is '-351 m/s'.

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The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy store
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A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}            (you has an mistake in the formula)

         \frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \  10^{6}

         \frac{1}{C_{eq1}} = 0.1   10⁶

         C_{eq1} = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          \frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}

          \frac{1}{C_{eq2} }  = (\frac{1}{15} + \frac{1}{30} +   \frac{1}{10} ) \ 10^6

          \frac{1}{C_{eq2} } = 0.2 10⁶

          C_{eq2} = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

           

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = \frac{Q^2}{2 C_3}

          U₃ =\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = \frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}

          U₄ = 0.5 J

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