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
[?] Pa
1 answer:
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;
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;
#SPJ1
You might be interested in
The range of the projectile is 188 m
Explanation:
The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The path of a projectile is the combination of these two motions: see figure in attachment.
In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.
We have:
t = 5.0 s (time of fligth of the projectile)
and the horizontal velocity is constant, and it is given by
where
is the initial velocity
is the angle of projection
Substituting,
And therefore, the range of the projectile is:
Learn more about projectile motion:
#LearnwithBrainly
Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
Then, the time of the second phase will be:
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
So, the total distance covered by this object in meters will be the sum of all the distances we found: