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Natasha_Volkova [10]
3 years ago
6

A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an

d 30 kg respectively. What is the normal force between the ground and one leg of the table?
Physics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

Therefore, rhe normal force is equivalent to 1,373.4 N

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Answer:

B. 1500 kg*m/s

Explanation:

Momentum p = m* v

In any type of collision, the total momentum is preserved!

The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.

p1 + p2 =

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Substitute the given numbers:

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1500 kg*m/s, which is answer B.

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