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miss Akunina [59]
4 years ago
6

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward a

cceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 64 m and acquired a velocity of 60 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The time interval (in s), during which the rocket engine provides upward acceleration, is closest to: ? .
Physics
1 answer:
Dvinal [7]4 years ago
4 0

Answer:

2.13 s

Explanation:

Hi!

At t = 0s the rocket is at rest in its platform, so the intial speed is zero. I f the acceleration is A, then the height Y, and the speed V are:

Y=\frac{A}{2}t^2

V=At

We nedd to find time T during  which the rocket engine provides upward acceleration. We know that:

64\;m=\frac{A}{2}T^2\\ 60\frac{m}{s} =AT\\

With these 2 equations we can find A and T (dropping units for simplicity):

A=\frac{60}{T} \\64 =\frac{30}{T} T^2=30T\\T=\frac{64}{30}\approx 2.13\;s

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Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

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Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

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b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
The tires of a car make 77 revolutions as the car reduces its speed uniformly from 95.0 km/h to 65.0 km/h. The tires have a diam
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Answer:

Explanation:

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65 km/hr = 18.06 m/s

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77(0.9π) = 69.3π m

v² = u² + 2as

a = (v² - u²) / 2s

a = (18.06² - 26.39²) / (2(69.3π))

a = -0.85 m/s²

s = (v² - u²) / 2a

s = (0² - 26.39²) / 2(-0.85)

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3 years ago
Which of the following would likely be the best reflector of heat energy?
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Answer: A, a polished pale yellow surface would likely be the best reflector of heat energy. 
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Answer:

Part a)

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As we know that the negative charge will experience the force due to some other charge below it

the force is given as

F = 0.700 N

now we know that

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now plug in all data

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So here the force on the charge below it will be same in magnitude but opposite in direction

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BlackZzzverrR [31]

Answer:

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<u>Vf = 0.0108 m³</u>

6 0
3 years ago
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