Ask question

Ask question

All categories

LuckyWell [14K]

3 years ago

13

Why do astronomers use spectroscopes to analyze light from distant objects?

1 answer:

RoseWind [281]3 years ago

7 0

Spectroscopy — the use of light from a distant object to work out the object is made of — could be the single-most powerful tool astronomers use, says Professor Fred Watson from the Australian Astronomical Observatory. ... "It lets you see the chemicals being absorbed or emitted by the light source.

You might be interested in

What is the relationship between the frequency and the pitch of a sound?

andrezito [222]

Choice-C is the correct one.

3 0

3 years ago

Read 2 more answers

Cindy runs 2 kilometers every morning. She takes 2 minutes for the first 250 meters, 4 minutes for the next 1,000 meters, 1 minu

PIT_PIT [208]

Answer:

200 m / min

Explanation:

Total distance = 2000 m

Total time = 2+4+1+3 = 10 minutes

Average speed = 2000 m / 10 min = 200 m/min

7 0

1 year ago

When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

Nadusha1986 [10]

Potential energy + kinetic energy = constant at every moment in time

At the highest point:

potential energy is at its maximum
kinetic energy is zero

8 0

3 years ago

Read 2 more answers

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

b)

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago