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Damm [24]
3 years ago
15

How much pressure is applied to the ground

Physics
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

Pressure applied by the man= 285103.125 Pa  or 41.35 lb/in^{2}

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e.  Pressure=\frac{Force}{Area}

Now, Force= mg

where, m = mass of the body(man) = 93 kg

g = acceleration due to gravity of Earth = 9.81 m/{s^{2}}

Area covered is equal to the area of both stilts(a man generally stands on two feet)

therefore Area=2(0.04)^{2} m^{2}

and putting in the values, we get,

Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}

Now we need to convert to our required units:

1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}

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If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth
Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

                                   F_b = Y * V_timber

                                   F_b = 9810*0.15*0.15*x

                                   F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

                                   (M)_p = 0

                                   W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

                                   670*0.5*3.6 - x^2 * 0.5*226.7 = 0

                                   x^2 = 1206 / 113.35

                                   x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

                                   sin(Q) = y / x

                                   sin(Q) = 2.1 / 3.26

                                   Q = sin^-1 (0.6441718)

                                   Q = 40.1 degrees

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A nanometer is a unit of mass, whereas a nanosecond is a unit of time. Question 1 options: True False
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Answer:

True.

Explanation:

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. To convert 1.3 hours to minute, you would multiply by 1 h / 60 min. Kilometer is a unit of length, whereas kilogram is a unit of mass. True.

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You want to place a mirror at a blind turn on the staircase in your house. Which would be best suited for this purpose?
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Which property is closely related to its temperature
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Answer:

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Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
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