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3 years ago

The table below shows the right ascensions of two stars at a location.

2 answers:

7 0

Star 1 - 4 hours right ascension
Star 2 - 3 hours right ascension
Subtracting hours right ascension
4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
star 1 will rise 1 hour before star 2

PtichkaEL [24]3 years ago

6 0

Answer:

The table below shows the right ascensions of two stars at a location.

Name of star Right ascension (in hours)

Star 1 1

Star 2 3

What is the degree difference between the two stars that cross the meridian at the location?

30 degrees

20 degrees

35 degrees

15 degrees

Explanation:

PLS ANSWER MINE TOO

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Which of the following is MOST useful to scientists in measuring the size of asteroids?

Alenkasestr [34]

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

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3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

3 years ago

A meteoroid is moving towards a planet. It has mass m = 0.78×109 kg and speed v1 = 4.1×107 m/s at distance R1 = 2.8×107 m from t

wariber [46]

Answer:

$PE=81.755\, J$

Explanation:

Given that:

mass of meteoroid, $m=7.8\times 10^8 \,kg$

radial distance from the center of the planet, $R= 2.8\times 10^7 m$

mass of the planet, $M=4.4\times 10^{25}\, kg$

<u>For gravitational potential energy we have:</u>

$PE=G\frac{M.m}{R}$

substituting the respective values:

$PE=6.67\times 10^{-11}\times \frac{4.4\times 10^{25}\times 7.8\times 10^8}{2.8\times 10^7}$

$PE=81.755\, J$

5 0

3 years ago

A man pushes a 10kg box with a constant acceleration of 5m/s2. What force is applied to the box?

drek231 [11]

FORMULA

$\boxed {F = m \times a}$

F = force
m = mass
a = acceleration

Using the formula

$F = 10 \times 5$

Multiply

$\boxed {\textsf {F = 10 N}}$

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3 years ago

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Sarah is looking for a way to move her boat across the water without touching it directly. She included paperclips in the boat d

Zigmanuir [339]

Answer:

When there is no detergent in the water, you'll achieve a floating paper clip!

Explanation:

3 0

3 years ago

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