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alukav5142 [94]
3 years ago
15

A space expedition discovers a planetary system consisting of a massive star and several spherical planets. The planets all have

the same uniform mass density. The orbit of each planet is circular.
In the observed planetary system, Planet A orbits the central star at the distance of 2R and takes T hours to complete one revolution around the star. Planet B orbits the central star at the distance of R. Which of the following expressions is correct for the number of hours it takes Planet B to complete one revolution around the star?

a. 1/â8T
b. 1/2T
c. 1/3â4T
d. 2T
e. â8T
Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

T/√8

Explanation:

From Kepler's law, T² ∝ R³ where T = period of planet and R = radius of planet.

For planet A, period = T and radius = 2R.

For planet B, period = T' and radius = R.

So, T²/R³ = k

So, T²/(2R)³ = T'²/R³

T'² = T²R³/(2R)³

T'² = T²/8

T' = T/√8

So, the number of hours it takes Planet B to complete one revolution around the star is T/√8

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The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
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Answer:

The answer is "60.74^{\circ}".

Explanation:

Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\

5 0
3 years ago
how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy
Blababa [14]
This question is probably referring to heat energy transferring from the car to its surroundings.
4 0
4 years ago
Science: Which statement is true about star A
Elza [17]

Answer:

no clue

Explanation:

6 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
3 years ago
A solenoid having N turns and carrying a current of 2.000 A has a length of 34 00 cm. If the magnitude of the magnetic field gen
umka21 [38]

Answer:

B) 1218

Explanation:

N = Total number of turns in the solenoid

L = length of the solenoid = 34.00 cm = 0.34 m

B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T

i = current carried by the solenoid = 2.000 A

Magnetic field at the center of the solenoid is given as

B = \frac{\mu _{o}N i}{L}

9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}

N = 1218

3 0
4 years ago
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