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3 years ago

If the velocity of an object changed from 30 m/s to 60 m/s over a period of 10 seconds what would the average acceleration be ?

Physics

1 answer:

Anarel [89]3 years ago

5 0

3 meters per second squared (3 m/s2)

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The sun, moon, planets and stars rise in the ____ and set in the ____.

Masja [62]

The sun, moon, planets and stars rise in the EAST and set in the WEST.

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3 years ago

A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in

Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

for the first one: 

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

for the second one :

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

a=20/6

a=3.33

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3 years ago

Read 2 more answers

How does the force of gravity between two bodies change when the distance between them doubles?

polet [3.4K]

Answer:

If the distance is doubled, the force of gravity between the two bodies is one-fourth as strong as before

Explanation:

The force of gravity between two bodies depends on the mass and distance. But we will focus on distance since that's what the question asks

Therefore, the force of gravity decreases as distance between the bodies increases.

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3 years ago

2. Turn off the Parallel line and turn on the Line through focal point. Move the light bulb around. What do you notice about the

MArishka [77]

Answer:

The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.

Explanation:

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3 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

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3 years ago