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Goshia [24]
2 years ago
5

A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the

air is stationary. What is the DIFFERENCE IN PRESSURE between the inside and outside?
[?] Pa​

Physics
1 answer:
Mariana [72]2 years ago
3 0

The difference in the pressure between the inside and outside will be 369.36 N/m²

<h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

\rm  \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\  P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\  \triangle p=369.36 \ N/m^2

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ1

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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
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Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

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7 0
3 years ago
A football player with a mass of 85 kg wears a uniform and helmet that have a mass of 4.5 kg. The football player moves at 2.1 m
DochEvi [55]
The answer would be 187.95 kg.m/s.

To get the momentum, all you have to do is multiply the mass of the moving object by the velocity. 

p = mv

Where:
P = momentum
m = mass
v = velocity

Not the question is asking what is the total momentum of the football player and uniform. So we need to first get the combined mass of the football player and the uniform. 

Mass of football player = 85.0 kg
Mass of the uniform     = <u>  4.5 kg</u>
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So now we have the mass. So let us get the momentum of the combined masses. 

p = mv
   = (89.5kg)(2.1m/s)
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melamori03 [73]
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