Answer:
The velocity is 
The pressure is 
Explanation:
From the question we are told that
The speed at which water is travelling through is 
The pressure is 
The diameter of the new pipe is
Where D is the diameter of first pipe
According to the principal of continuity we have that
Now
is the area of the first pipe which is mathematically represented as

and
is the area of the second pipe which is mathematically represented as

Recall
![A_2 = \pi \frac{[ D^2]}{4 *4}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20%20%5Cfrac%7B%5B%20D%5E2%5D%7D%7B4%20%2A4%7D)

So 
substituting value

According to Bernoulli's equation we have that

substituting values


The right answer is. This kind of phase change--liquid<span> to </span>gas<span>--is </span>called<span> evaporation or vaporization. Water vapor can in </span>turn<span> be cooled to form </span>liquid<span> water. This kind of phase change--from</span>gas<span> to </span>liquid<span>--is referred to as condensation.</span>
Explanation:
ultraviolet radiation
Bombarded with energy
They stored the seeds in a single layer on the outside of the ISS behind a special kind of glass that let in ultraviolet radiation only at wavelengths between 110 and 400 nanometers
Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
- W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
= K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} =
+ U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
Answer:
18 ohms
Explanation:
V = I(R1 + R2)
5V = (0.167A)(12 ohms + R2)
Solving for R2
R2 = 18 ohms