Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
P = F / A
Where the area of a circle is
A = π r²
E radius is half the diameter
r = d / 2
A = π d² / 4
We replace
P = F 4 / π d²2
P₁ = 397 4 /π 0.058²
P₁ = 1.50 10⁵ Pa
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
P₁ = P₂ + ½ ρ v₂²
v₂ = √ (P1-P2) 2 / ρ
v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
v₂ = 97.4 m / s
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:
We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
![|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}](https://tex.z-dn.net/?f=%7Cv_%7Bhawk%7D%7C%3D%5Csqrt%5B%5D%7B6%5E2%2B2%5E2%7D%3D%5Csqrt%7B40%7D)
≈
And the angle with respect to the east should be with:
A light wave that hits the surface of a pool gets refracted and gives us an apparent image of the surface of the pool, following the concepts of refraction.
<u>Explanation:</u>
Let’s recall the concept of refraction when a light wave passes from medium of rarer to denser. There is a change in the speed of light while travelling from medium of rarer to denser.
There can be a change in the direction as well. This property is known as “Refraction” and the best example to see refraction is watching the surface of a clean pond, lake or pool.
When the light travels from a rarer medium (air) to a denser medium (water), it changes its angle of direction and gets refracted and hit to our eye lenses. With this, we see the surface of the pool at a changed angle and it seems to be a bit shallow than its original depth.
An antibaryon composed of two antiup quarks
and one antidown quark would have a charge of (2) −1e.
The answer is 3) 3.00 m/s2
