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Grace [21]
3 years ago
15

S Problem Set 2.) 6.4 x 109 nm to cm

Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

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Early telescope makers compensated for chromatic aberration by:
svp [43]
Increasing the focal lenght
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3 years ago
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If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is
Law Incorporation [45]

Answer:

3

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

3 0
2 years ago
Please, someone help me I'm begging yall 1. it's estimated that 1 kg of body fat will provide 3.8 * 10^7 J of energy. A 67 kg mo
Leto [7]

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed

7 0
3 years ago
Why doesn't the earth block out the light between the sun and the moon during a normal month
almond37 [142]

Answer:

Explanation:

6 0
3 years ago
Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth?
aniked [119]

Answer:

Low satellite has high orbital velocity

Explanation:

let v be the orbital speed of the satellite orbiting at a height h is given by

v=\sqrt{\frac{GM}{R+h}}

where, M be the mass of planet, r be the radius of planet and h be the height of planet from the surface of planet.

here we observe that more be the height lesser be the orbital velocity.

So, a satellite which is at low height has high orbital velocity.

3 0
4 years ago
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