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3 years ago

What type of telescope can be used routinely on the surface of the earth during the day?

Physics

1 answer:

nlexa [21]3 years ago

5 0

Answer:

Radio telescopes .

Explanation:

They have a modest size that can't make out as much detail (have a lower resolution) than visible light telescopes

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A ball is dropped off the side of a bridge. after 1.55 s, how far has it fallen

givi [52]

Answer:

38.64 feet

Explanation:

x=x0 + vx0t + 1/2axt2

x= 0 + 0 + 1/2 X 32.17 ft/sec2 X 1.55 sec2

x = 38.64 feet

7 0

3 years ago

What happens to the atoms that make up hydrogen fuel as it burns?

ale4655 [162]

Answer:

Get turned Into Water.

Explanation:

Combustion of Hydrogen involves combining oxygen and hydrogen essentially so when oxygen and hydrogen combine water is produced, following chemical equation describes this process.

$H_{2} +O_{2}=2(H_{2}O)$ .

Resulting product is two molecules of water.

4 0

4 years ago

a rifle is fired and recoils when the bullet leaves the gun. this is an example of newton's 3rd law. the force on the bullet is.

Alex_Xolod [135]

The force of the bullet is the same.

3 0

3 years ago

Read 2 more answers

A paper airplane is thrown to the north from the top of a cliff with a speed of 2 km/h. The paper airplane lands 15 minutes (0.2

Luden [163]

If you neglect air resistance, then you can solve for the horizontal distance using the formula:

dx=vixt

Where:
dx = horizontal distance
vix = initial horizontal velocity
t = time in flight

Now you can see that you have all the given you need.
vix= 2km/h
t = 0.25 hours

We use hours because you need the units to be the same. So just plug that in your equation and you will have your answer:

dx = vixt
=(2km/h)(0.25h)
=0.5km

This means that the answer to your question is B.

5 0

3 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago