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umka2103 [35]
3 years ago
5

Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.001

33 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Physics
1 answer:
Vlad1618 [11]3 years ago
3 0

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

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A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to W=mg (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to B=m_w g (m_w=mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

W'=W-B

which can be rewritten as

m'g = mg-m_w g

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

m_w = m-m'=540 g-342 g=198 g

(b) 1.98\cdot 10^{-4} m^3

If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

where

m_w = 198 g = 0.198 kg is the mass of the water displaced

\rho_w = 1000 kg/m^3 is the density of the water

Substituting,

V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3

And so this is also the volume of the rock.

(c) 2727 kg/m^3

The average density of the rock is given by

\rho = \frac{m}{V}

where

m = 540 g = 0.540 kg is the mass of the rock

V=1.98\cdot 10^{-4} m^3 is its volume

Substituting into the equation, we find

\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

3 0
3 years ago
An electric field of intensity 3.80 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
lakkis [162]

Answer: 1.55 x 10⁴ Nm²c^-1

Explanation: The electric flux, electric field intensity and area are related by the formulae below.

Φ= EAcosθ,

Where Φ= electric flux (Nm²c^-1)

E =electric field intensity (N/m²)

A = Area (m²)

θ= this is angle between the planar area and the magnetic flux

For our question E=3.80KN/c= 3800 N/c

A= 0.700 x 0.350= 0.245m²

θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).

Hence Φ= 3800 x 0.245 x cos(0)

= 3800 x 0.245 x 1 (value of cos 0° =1)

= 1.55 x 10⁴ Nm²c^-1

Thus the electric field is 1.55 x 10⁴ Nm²c^-1

4 0
3 years ago
An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the
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Answer:

9.25 x 10^-4 Nm

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number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

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Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

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