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disa [49]
2 years ago
11

How much do you do when you stand by carrying 50 kg load bricks? write the reason​

Physics
1 answer:
yanalaym [24]2 years ago
6 0

Answer:

No work is done yet

Explanation:

Given data

Weight of the load=50kg

Distance= 0 meter ------------Standing on one point

Normally, work is said to be done when we apply a force/carry a load through a distance, since the person is standing and not moving through a distance, no work is done

Hence

Work done= 50*0= 0 joules

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1.loudness

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Juan compró un carro que dicen que es muy rápido. Cuándo lo probó recorrió una distancia de 4500m en tan solo 5 min. ¿Qué veloci
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Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

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What interests you the most about Social Psychology? What do you think is one of the most aggravating issues that our society fa
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6 0
3 years ago
Three identical very dense masses of 3500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,
larisa86 [58]

Answer:

<em>A) 7.37 x 10^-4 N</em>

<em>B) The resultant force will be towards the -x axis</em>

Explanation:

The three masses have mass = 3500 kg

For the force of attraction between the mass at the origin and the mass -100 cm away:

distance r = 100 cm = 1 m

gravitational constant G= 6.67×10^−11 N⋅m^2/kg^2

Gravitational force of attraction F_{g} = \frac{Gm^{2} }{r^{2} }

where G is the gravitational constant

m is the mass of each of the masses

r is the distance apart = 1 m

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{1^{2} } = 8.17 x 10^-4 N

For the force of attraction between the mass at the origin and the mass 320 cm away

distance r = 320 cm = 3.2 m

F_{g} = \frac{Gm^{2} }{r^{2} }

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{3.2^{2} } = 7.98 x 10^-5 N

Resultant force = (8.17 x 10^-4 N) - (7.98 x 10^-5 N) = <em>7.37 x 10^-4 N</em>

<em></em>

<em>B) The resultant force will be towards the -x axis</em>

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3 years ago
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