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prisoha [69]
3 years ago
8

What is a charge-coupled device (CCD), and how is it used in astronomy?

Physics
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

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7. Starting at rest, a car accelerates at 5.5m/s/s for 12s. What is its
Musya8 [376]

Answer:

66 m/s

Explanation:

v=u+at

= 0 + 5.5 * 12

= 66 m/s

6 0
3 years ago
A point source emits 25.9 W of sound isotropically. A small microphone intercepts the sound in an area of 0.242 cm2, 590 m from
AnnZ [28]

The solution is in the attachment

7 0
4 years ago
When temperature decreases but actual water vapor content remains the same, what happens to relative humidity?
ruslelena [56]

Answer:

d. Relative humidity increases.

Explanation:

The expression of relative humidity in terms of absolute humidity, absolute pressure and saturation pression at measured temperature is:

\phi = \frac{\omega \cdot P}{(0.622+\omega)\cdot P_{sat}}

When temperature decreases, the saturation pressure decreases also and, consequently, relative humidity increases. Therefore, the right answer is option D.

8 0
3 years ago
Read 2 more answers
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
The graph represents velocity over time...
Nady [450]

Answer:

where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please

8 0
2 years ago
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