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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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belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

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if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = \sqrt{(0.38m)^2+(0.38m)^2}

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

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4 0
4 years ago
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Viktor [21]

Answer:

λ = 8.88 x 10⁻⁷ m = 888 nm

Explanation:

The energy band gap is given as:

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c = speed of light = 3 x 10⁸ m/s

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Therefore,

2.24 x 10⁻¹⁹ J = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.24 x 10⁻¹⁹ J)

<u>λ = 8.88 x 10⁻⁷ m = 888 nm</u>

7 0
3 years ago
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D = \sqrt{N^2 + W^2}

D = 4.93 Km

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7 0
3 years ago
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Answer:

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