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3 years ago

A race car accelerates from 0.00m/s to 40.0m/s with a displacement of 50.0m. What was the cars

Physics

1 answer:

Vladimir79 [104]3 years ago

7 0

Answer:

Explanation:

Using the equation of motion v² = u²+2as

v is the final velocity = 40m/s

u is the iniyail velocity 0m/s

a is the acceleration

s is the displacement

Substituting in the formula;

40² = 0²+2a(50)

1600 = 100a

Divide both sides by 100

100a/100 = 1600/100

a = 16

Hence the car acceleration is 16m/s²

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The mass of this astronaut on the moon is 84kg. If the astronaut is on the moon for 3 weeks then returns home, his mass will be

saul85 [17]

Answer:

84kg

Explanation:

If the mass of an astronaut in on the moon is 84kg on the moon. After the 3weeks in returning home, his weight will not change. The mass of the astronaut will remain the same which is 84kg. This is so because the acceleration due to gravity experienced by the body remains the same both in the earth and in moon.

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3 years ago

all of the following answer choices are ideas from daltons atomic theory. which one of them do we now know is not true?

dem82 [27]

Did you forget to include the options?

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If a 50kg boulder is 100m off the ground and a 25 kg boulder is 100 m off the ground, which has a

Flauer [41]

Answer:

Explanation:

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3 years ago

The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8

qwelly [4]

Answer: f = -12 cm

Explanation: <u>Combined</u> <u>lenses</u> is an array of simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

If there is a distance between the lenses, the focal length will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}$

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

$\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}$

$\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}$

$\frac{1}{f_{2}} =\frac{-2}{24}$

$f_{2}=$ -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

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3 years ago

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malfutka [58]

Answer:

The correct option is: d. Fiber-optic

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3 years ago