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Vinil7 [7]
2 years ago
11

The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.4 earth years.

Physics
1 answer:
Ad libitum [116K]2 years ago
3 0

Answer:

(a) Radius = 4.6 x 10^11 m

(b) speed = 16.96 km/s

Explanation:

Time period, T = 5.4 earth years

mass of sun, M = 1.989 x 10^30 kg

(a) Let the orbital radius is R.

use the formula of period

T^2 = \frac{4 \pi^2 R^3}{G M}\\\\\left ( 5.4\times 365\times 24\times 3600 \right )^2 = \frac{4\times3.14\times 3.14\times R^3}{6.67\times10^{-11}\times 1.989\times 10^{30}}\\\\R = 4.6\times 10^{11} m

(b) Let the speed is v.

v=\frac{2 \pi\times R}{T}\\\\v=\frac{2\times 3.14\times 4.6\times 10^{11}}{5.4\times 365\times 24\times 3600}\\\\v = 16963.6 m/s =16.96 km/s

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4 0
3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
Help me, Ive been stuck for 15 minutes
natita [175]
The answer is b) venus because the period rotation is also referred to as a day so if venus’ period of rotation is 243 [earth] days and its year is 225 [earth] days, then the period of rotation is longer.
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Through x rays is harmful for human health it is used in medical field why?
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It is strong enough to penetrate through flesh but not bone so we can see if there are fractures or breaks in our skeleton
6 0
2 years ago
Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number o
lys-0071 [83]

Answer:

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1(x, t) = Asin(kx - ωt)

y1(x, t) = 7sin(3x - 2.5t)

Wave 2 has a wave function:

y2(x, t) = Asin(kx - ωt + ϕ)

y2(x, t) = 7sin(3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y(x, t) = 7sin(3x - 2.5t) + 7sin(3x - 2.5t + π/12)

y(x, t) = 7[sin(3x - 2.5t) + sin(3x - 2.5t + π/12)]

Using trigonometric function:

sin(a) + sin(b) = 2cos[(a - b)/2]sin[(a + b)/2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y(x, t) = (2*7)[cos(π/24)sin(3x - 2.5t + π/24)]

Therefore, when x = 0.53cm and t = 2s,

y(x, t) = (2*7)[cos(π/24)sin{(3*0.53) - (2.5*2)+ π/24}]

y(x, t) = 14 * 0.9914 * 0.0713

y(x, t) = 0.99 m

The height of the resultant wave is 0.99cm

5 0
2 years ago
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