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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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Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

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= 0.64 m

Mechanical advantage,

= 0.16

As we know,

⇒ Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}

On putting the values, we get

⇒                         =\frac{0.64}{0.16}

⇒                         =4 \ m

4 0
3 years ago
Identical forces act for the same length of time on two different masses. The change in momentum of the smaller mass is
xenn [34]

Answer:

Equal to change in momentum of larger mass.

Explanation:

We are given that

Two difference  masses .

Force act on both masses  for the same length of time.

We have to find the change in momentum of the smaller mass.

Let M and m are two masses

M>m

We know that

Change in momentum for large mass=F\Delta t

Change in momentum for small mass=F\Delta t

Because Force and length of time are same for both masses .

Hence, the change in momentum of smaller mass is equal to change in momentum of larger mass.

7 0
3 years ago
Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i
Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done  = 379.5J

Height  = 173m

Unknown:

Amount of force exerted on the sled  = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

      Work done  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

     mg  = weight;

        Work done  = weight x h

           379.5 = weight  x  173

           weight  = \frac{379.5}{173}   = 2.2N

4 0
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The magnification is less than 1 what does it mean​
igomit [66]
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4 0
3 years ago
A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

\alpha = Angular Acceleration

Torque

\tau=F\times r

\tau=I\times \alpha

\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}

Angular velocity

\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s

Angular acceleration

\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2

I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2

F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0
3 years ago
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