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Nostrana [21]
3 years ago
15

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floo

r and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?
Physics
1 answer:
qaws [65]3 years ago
4 0

Answer:

Force, F=2.02\times 10^6\ N

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, P=1\ atm=101325\ Pa

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

P=\dfrac{F}{A}

F=P\times l\times b

F=101325\times 4\times 5

F=2.02\times 10^6\ N

So, the total force on the floor due to the air above the surface is 2.02\times 10^6\ N. Hence, this is the required solution.

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By how much would 1500j of heat energy raise the temperature of 0.50kg of aluminum​
valentinak56 [21]

pls what is the specific heat capacity of water

7 0
3 years ago
Why dont we put (-ve) sign when calculating charge in electrostatics ?​
Levart [38]

Answer:

The negative sign represents the flow of charge in an opposite direction relative to point of action.

5 0
3 years ago
A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
Readme [11.4K]

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

D=\sqrt{9^2+12^2} \\\\D=15\ km

For direction,

\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}

Hence, this is the required solution.

5 0
3 years ago
The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?
tensa zangetsu [6.8K]
M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse = <span> m(v - u)</span>
 
             = 150*(14 - 6)
           
             =  150*8 = 1200 kgm/s  or 1200 Ns<span> </span>
3 0
3 years ago
An amount of energy is added to ice, raising its temperature from -10°C to -5°C. A larger amount of energy is added to the sam
lara [203]

Answer:

The correct option is;

B) The specific heat of ice is less than that of water.

Explanation:

Here we have

Let the amount of energy added to the ice at -10 C to raise the temperature to -5 C be X J

Let the amount of energy added to the water at 15 C to raise the temperature to 20 C be Y J

We know that the heat required, ΔQ to raise the temperature of a substance is given by

ΔQ = m·c·Δθ

Where:

m = Mass of the substance

c = Specific heat capacity

Δθ = Temperature change

Since the mass of the ice and the water are the same, so also is the change in temperature, (-5 - (-10) = 5 and 20 - 15 = 5) we have

for m₂·c₂·Δθ₂ > m₁·c₁·Δθ₁

Where:

m₁, c₁, Δθ₁, is for the ice and m₂, c₂, Δθ₂ is for the water and

m₁ = m₂

Δθ₁ = Δθ₂

Therefore,

c₂ > c₁ = c₁ < c₂

That is the specific heat capacity of the ice is lesser than the specific heat capacity of the water.

4 0
3 years ago
Read 2 more answers
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