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nikitadnepr [17]
3 years ago
8

. An iron cube (ρ = 7860 kg/m3) having sides of 1 cm is submerged in open fresh water (ρ = 1000kg/m3) with its top surface at a

depth of 2.5 m. Considering that its top surface is held horizontal with respect to the water surface, determine the difference in pressure between the top and bottom surfaces of the cube?
Physics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

\Delta P = 98.07\,Pa

Explanation:

The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:

\Delta P = \rho_{w}\cdot g \cdot \Delta h

\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.01\,m)

\Delta P = 98.07\,Pa

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svetlana [45]

300

Explanation:

100 x 3 =300 simple and easy

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2 years ago
Objects that transmit some light and reflect/absorb the rest of the light, like stained glass, are
11111nata11111 [884]
You are correct it is translucent because opaque let’s no light through but transparent lets almost all light through
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3 years ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
Suppose that you and three classmates are discussing the design of a roller coaster. One says that each hill must be lower than
8_murik_8 [283]

Answer:

I would say that I agree with the one that said that each hill must be lower than the previous one and use the principle of conservation of energy to explain.

Explanation:

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However, due to the effect of gravity and frictional force between the track and the coaster, the energy of the coaster is gradually reduces, so it becomes a bit difficult for the coaster to move to the next hill of the same height. It is for this reason that each hill must be lower than the previous one, so that the coaster can overcome the next hill's height with its reduced energy until it loses all its energy and comes to a stop.

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3 years ago
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Answer:

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