Answer:
a) Weight of the rock out of the water = 16.37 N
b) Buoyancy force = 4.61 N
c) Mass of the water displaced = 0.47 kg
d) Weight of rock under water = 11.76 N
Explanation:
a) Mass of the rock out of the water = Volume x Density
Volume = 470 cm³
Density = 3.55 g/cm³
Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg
Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N
b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.
Volume = 470 cm³
Density of liquid = 1 g/cm³
c) Mass of the water displaced = Volume of body x Density of liquid
Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg
d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force
Weight of rock under water = 16.37 - 4.61 =11.76 N
Answer:
Explanation:
When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.
In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:
where
is the wavelength in a vacuum (air is a good approximation of vacuum)
n is the refractive index of the medium
In this problem:
is the original wavelength of the wave
n = 1.47 is the index of refraction of corn oil
Therefore, the wavelength of the electromagnetic wave in corn oil is:
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
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