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Sati [7]
3 years ago
10

How much must a man weigh (force) if the pressure he exerts while standing on one foot which has an area of 25 m^2 exerts a pres

sure of 5 pa?
Physics
1 answer:
andriy [413]3 years ago
7 0

Answer:

Weight, W = 125 N

Explanation:

Pressure, P = 5 Pa

Area, A = 25 m²

We need to find how much must a man weigh (force) if the pressure he exerts while standing on one foot which has an area of 25 m² exerts a pressure of 5 Pa. So,

Pressure = force/area

So,

F=P\times A\\\\F=5\times 25\\F=125\ N

So, the weight of the man is 125 N.

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The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
 We have then:
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 Then, the resulting net force is:
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 Then by definition:
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 Clearing the acceleration:
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 a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
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3 years ago
Which of the following elements would not be considered a nonmetal
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There are no nonmetals on the list you provided.
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What are the properties of the layers of the Earth atmosphere
pantera1 [17]

Answer:

The atmosphere has 4 layers: the troposphere that we live in near the surface of the earth; the stratosphere that houses the ozone layer; the mesosphere, a colder and lower density layer with about 0.1% of the atmosphere; and the thermosphere, the top layer, where the air is hot but very thin.

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3 years ago
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Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
If 317. 7 coulombs was used in an electrolytic cell, how many grams of zn would we expect to be plated from a zn2 solution onto
viva [34]

Answer: The amount of Zn plated from a solution is 1.07×10^{-1} grams.

Explanation: In this question, it is given that of 317.7 coulombs was used in an electrolytic cell, i.e,

The amount of charge = 317.7 coulombs

The amount of charge of 1 mole of electrons = 96485.3 coulombs   ( FARADAY’S CONSTANT )

Now,    the number of electrons = amount of charge given / charge of 1 mole of electrons

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Now, we know that the molar mass of zinc is 65.38gm/mole,

          Number of moles = mass in grams / molar mass

          Mass in grams = number of moles × molar mass

          Mass in grams =  (1.645×10^{-3}) × 65.38  =  0.1075 gm.

Conclusion: It concludes that the amount of Zn plated from a 1.07×10^{-1}  solution is  gm.

3 0
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