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allochka39001 [22]

3 years ago

10

The light from one end of the Milky Way galaxy takes about 150,000 years to travel to the other end. Which of these statements i

s correct about the Milky Way galaxy?
A: The distance between the two ends of the galaxy is 150,000 light years.

B: It would take 150,000 years to photograph one end of the galaxy from the other end.

C: Both ends of the galaxy would meet if they move a distance of 75,000 kilometers each.

D: Both ends of the galaxy would meet if they move towards each other at the rate of 75,000 kilometers per hour.

1 answer:

Nataliya [291]3 years ago

8 0

<span>A: The distance between the two ends of the galaxy is 150,000 light years.</span>

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Which of the following most likely happens when the temperature of a chemical reaction is increased?

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Molecules Speed Up because the other three options only occur when the temperature is decreased.

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

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Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

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