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2 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?

1 answer:

qaws [65]2 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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Find the net force for <br> 10 N<br> 10 N<br> 25degree<br> 5N<br> 5N

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Answer:

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Explanation:

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2 years ago

A 15 N force and a 45 N force act on in object in opposite directions. What is the net force on the object?

vovikov84 [41]

Answer:

30N in the direction the 45N acts.

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Fnet = F1 + F2 (the vector sum of the forces)

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Fnet = 45 - 15

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2 years ago

A bird can fly 25 km/h how long does it take to fly 3.5km

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2 years ago

The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100

Tema [17]

Answer:

drag force $F_D = 1.5 \ N$

Velocity (V) = 40.169 m/s

Explanation:

The drag force $F_D$ is given by the formula:

$F_D = C_D * \frac{1}{2}* \rho * V^2*A$

where:

$C_D$ = drag coefficient depending on the Reynolds number

Reynolds number Re = $\frac{\rho *V*D}{ \mu}$

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air $\rho$ = 1.1845 kg/m³

viscosity of fluid or air $\mu$ = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

$Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5$

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

$C_D$ = 0.3

Projected Area A = $\frac{\pi D^2}{4}$

A = $\frac{\pi 0.074^2}{4}$

A = 0.0043 m²

Finally, drag force is then calculated as ;

$F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N$

b)

$- F_D = ma$

since acceleration a = $\frac{dV}{dt}$

Then;

$-F_D = m \frac{dV}{dt}$

Also;

velocity (V) = $\frac{dx}{dt}$

Then;

$- F_D = \frac{md_2x}{dt^2}$

$\frac{d_2x}{dt^2} = \frac {- F_D}{m}$

$F_D = 1.5 \ N\\m = 0.15 \ kg$

Then;

$\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}$

$\frac{d_2x}{dt^2} =- 10$

Integrating the above equation ; we have :

$\frac{dx}{dt}= - 10 t + C\\$

when time (t) = 0 ; then $\frac{dx}{dt}= V = 44.3$

44.3 = - 10 × 0 + C

C = 44.3

$\frac{dx}{dt}= V = -10 t + 44.3$

Time (t) =

$\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s$

∴ Velocity ; $\frac{dx}{dt}= V = - 10t +44.3$

$\frac{dx}{dt}= V = - 10(0.413 s) +44.3$

Velocity (V) = 40.169 m/s

6 0

3 years ago