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fgiga [73]
3 years ago
13

Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona

de 60 kg de masa que se encuentre sobre su superficie?
Physics
1 answer:
erica [24]3 years ago
4 0

Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

Mass of the person, m = 60 kg

The acceleration due to gravity on the surface of gravity is 3.7 m/s²

We need to find the weight of a person on the surface of Mercury.

Weight of an object is given by :

W = mg

So,

W = 60 kg × 3.7 m/s²

W = 222 N

Hence, the person will weigh 2222 N on the surface of Moon.

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1. Na jaką wysokość wzniesie się piłka kopnięta pionowo w górę z prędkością 20 m/s ? 2. Piłka o masie 20 dag spada z wysokości 1
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its gonn abe 23

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I need help with these four problems.. anyone??
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3 years ago
A cadillac escalade has a mass of 2 569.6 kg, if it accelerate at 4.65m/s2 what is the net force of the car
anygoal [31]
Force is equal to mass multiplied by acceleration, therefore 
F=ma
m=2569.6 kg
a=4.65m/s^2
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3 0
3 years ago
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat
RSB [31]

Answer:

Explanation:

The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.

It also does say that each chair weighs 250 kg, and as such the load is

M = 50 * 250

M = 12500.

Taking into consideration, the initial and final heights, we have

h1 = 0, h2 = 200 m

The work needed to raise the chairs,

W = mgh, where h = h2 - h1

W = 12500 * 9.81 * (200 - 0)

W = 2.54*10^7 J

The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be

t = 1/10 = 0.1 h or say, 360 s

The power needed thus, is

P = W/t

P = 2.54*10^7 / 360

P = 68125 W, or 68 kW

Initial velocity, u = 0 m/s

Final velocity, v = 10 km/h = 2.78 m/s

Startup time, t is 17 s

Acceleration during the startup then is

a = (v - u)/t

a = 2.78/17

a = 0.163 m/s²

The power needed for the acceleration is

P = ½m [(v² - u²)/t]

P = ½ * 12500 * [2.78²/17]

P = 6250 * 0.455

P = 2844 W

3 0
3 years ago
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