4 years ago

What is the magnitude of g at a height above Earth’s surface where free- fall acceleration equals 6.5 m/s2 ?

1 answer:

6 0

Well, the "magnitude of g" in a place is always described as the
free-fall acceleration there. So your ' 6.5 m/s² ' is the answer to
the question.

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If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left

viva [34]

Answer:

Focal length, f = 16 cm

Explanation:

Image distance, v = 24 cm

Object distance, u = -48 cm

We need to find the focal length of the lens. It can be determined using the lens formula as :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{24\ cm}-\dfrac{1}{(-48\ cm)}=\dfrac{1}{f}$

f = 16 cm

So, the focal length of the converging lens is 16 cm. Hence, this is the required solution.

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mina [271]

Answer:

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3 years ago

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riadik2000 [5.3K]

I believe the best option to go with will be option C. electrons are negatively charged

4 0

4 years ago

If a car is traveling forward at 15 M/S, how fast will be going in 1.2 seconds if the acceleration is 10 M/s2?

GuDViN [60]

Answer:

Vf=27m/s

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Vf=Vi+at

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6 0

3 years ago