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3 years ago

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scientist can measure the depths of crayers on the moon by looking at pictures of the shadows. the length of the shadow cast by

edge is 500 meters. the angle of elevation of the rays of the Sun is 55 degrees estimate the depth of the crater

1 answer:

OLga [1]3 years ago

7 0

Answer: $d=714 meters$

Explanation:

Tita: $tita= 55 degrees$

depth of the crater: d= ?

length of the shadow cast by edge: $l= 500 meters$

$opposite/adjacent = tan tita$

$d/500=tan55degrees$

cross multiply

$d=tan55 degrees *500$

$d=1.4281*500$

$d=714 meters$

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Downwelling is the process that moves cold, dense water from the ocean surface to the seafloor near the polar regions. how can d

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Explanation:

Downwelling is the process where cold and heavy dense water moves down into the ocean floor and warm light dense water rises to the surface. As a result of downwelling, the water high dense water which rises to the water surface brings the oxygen rich water to the surface for the marine animals to breathe properly. Also when the ocean surface water becomes little warmer it becomes a little comfortable for the marine animals to survive in this severely cold climatic conditions at polar reasons.

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4 years ago

You get up in the morning, get dressed, eat breakfast, walk to the bus stop, and ride to school. List three different energy tra

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3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s

3 0

4 years ago

Read 2 more answers

Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble

stich3 [128]

From the calculations, the final momentum of B is 8.16 m/s

<h3>What is conservation of linear momentum?</h3>

According to the principle of the conservation of linear momentum, the momentum before collision is equal to the total momentum after collision.

This implies that;

MaUa + MbUb = MaVa + MaVa

Substituting values;

(0.08 kg * 0.5 m/s) + (0.05 kg * 0 m/s) = (0.08 kg * −0.1 m/s) + (0.05 kg * v)

0.4 = -0.008 + 0.05v

v = 8.16 m/s

Learn more about more about momentum: brainly.com/question/24030570:

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2 years ago

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

$r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m$

So, we can re-arrange equation (1) to find the orbital period:

$T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s$

Dividing by 60, we get that this time corresponds to 94 minutes.

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3 years ago

Read 2 more answers