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Sati [7]
3 years ago
15

What type of visible light spectrum does the sun produce?

Physics
1 answer:
VLD [36.1K]3 years ago
4 0
UV Rays. - yung bandz
                       


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how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum
dangina [55]

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

6 0
2 years ago
Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during
suter [353]

Answer: Option <em>a.</em>

Explanation:

Kepler's 2nd law of planetary motion states:

<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

3 0
3 years ago
A 650 × 10–4 F capacitor stores 24 × 10–3 of charge. What is the potential difference between the plates?
olasank [31]

Answer:

0.369 V

<u>Explanation:</u>

Given :

Capacitance ( c ) = 650 × 10–4 F

Charge ( q ) = 24 × 10–3 C

We are asked to find potential difference ( v )!

We know:

q = c v

= > v = q / c

Putting values here we get:

= > v = ( 24 × 10–3 ) / ( 650 × 10–4 ) V

= > v = 240 / 650 V

= > v = 24 / 65 V

= > v = 0.369 V

Therefore, potential difference between the plates is 0.369 V

8 0
3 years ago
The energy necessary to enable a reaction to occur is called the activation energy.
12345 [234]
True. In order for a chemical reaction between elements/compounds, the atoms within the chemicals must have sufficient energy in order to be able create a reaction.
5 0
3 years ago
5. A 905 kg test car travels around a 3.04 km circular track. If the magnitude of the centripetal force is 2100 N. What is the c
balu736 [363]

Answer:

Explanation:

The equation for centripetal force is

F=\frac{mv^2}{r}. We have all the values we need except for the radius. We have the circumference of the circle, though, so we will solve for the radius using that and the fact that C = 2πr:

3.04 = 2(3.1415)r and

r = .484 m, to the correct number of sig fig's.

Now that we have everything we need and isolating the v NOT squared:

v=\sqrt{\frac{rF}{m} } and filling in:

v=\sqrt{\frac{(.484)(2100)}{905} } . This answer will need 2 sig fig's since 2100 has 2 sig fig's in it. That means that the velocity of the test car is

1.1 m/sec

8 0
2 years ago
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