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kondaur [170]
1 year ago
14

the deepest point in the sea is 100m below sea level .what arr the water pressure at this depth and the total pressure due to wa

ter and atmosphere ?​
Physics
1 answer:
fomenos1 year ago
4 0

The water pressure at this depth and the total pressure due to water and atmosphere are  10.3 x 10⁵ Pa and 11.31× 10⁵ Pa.

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Atmospheric pressure, Patm =1.01×10⁵ Pa

Density of water, ρ=1030 kg/m³

Depth, h=100 m

Pressure =ρgh

P = 1030×10×100

P = 10.3 x 10⁵ Pa.

Total pressure, P=Po +ρgh

P=1.01×10⁵ + 1030×10×100

P=11.31× 10⁵ Pa

Hence, total pressure is 11.31× 10⁵ Pa.

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

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4 0
3 years ago
Two cylinders with the same mass density rC = 713 kg / m3 are floating in a container of water (with mass density rW = 1025 kg /
cluponka [151]

Answer:

\dfrac{h_2}{h_1}=\dfrac{1}{2}

Explanation:

Lets take h is height of cylinder immersed in the water

We know that for floating body

h=\dfrac{\rho_cL}{\rho_l}

Where \rho_c density of cylinder

\rho_l density of water

For both cylinder fluid is same also density of cylinders are also same

So

\dfrac{h_1}{L_1}=\dfrac{h_2}{L_2}

\dfrac{h_1}{h_2}=\dfrac{L_1}{L_2}

\dfrac{h_1}{h_2}=\dfrac{20}{10}

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3 0
2 years ago
Dose sound travel faster in a warm room or a cold room? explain your answer
dusya [7]

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is the ratio of the specific heats

R is the gas constant

T is the temperature of the medium

We know that the temperature of the warm room is more as compared to the cold room.

So, it is clear that the sound travel faster in a warm room. The particles move faster when the temperature is high.            

8 0
2 years ago
6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating
Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P  =  55 kW

heating value, H = 44,000 kJ/kg

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Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

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Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

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