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Pavel [41]
3 years ago
9

A city uses a water tower to store water for times of high demand. When demand is light, water is pumped into the tower. When de

mand is heavy, water can flow from the tower with- out overwhelming the pumps. To provide water at a typical 350 kPa gauge pressure, how tall must the tower be?
Physics
1 answer:
love history [14]3 years ago
6 0

Answer:

The height of the tower will be 35.714 m

Explanation:

We have given gauge pressure P=350kPa=350\times 10^3Pa

Density of water \rho =1000kg/m^3

We have to find the height of the tower h

We know that gauge pressure is given by P=\rho gh

350\times 10^3=100 0\times 9.8\times h

h=35.714m

So the height of the tower will be 35.714 m

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What occurs during stage one sleep
Mekhanik [1.2K]

Answer:

Stage one of sleep, also known as the transitional phase, occurs when one finds themselves floating in and out of consciousness. During this NREM stage, you may be partially awake while your mind begins to drift off. This period of drowsiness eventually leads to a light sleep

Explanation:

i found it on google

3 0
3 years ago
You run 100 meters in 50 seconds. Calculate your speed, label with m/s
NemiM [27]

Answer:

2 m/s

Explanation:

speed = distance/time

s = 100/50 = 2

4 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
2 years ago
Based on the law of conservation of energy, which statement is false?
alexgriva [62]

I think it’s Energy is lost when machines don’t work right.

8 0
2 years ago
A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the
Norma-Jean [14]

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

t = time interval = 4.5 s

Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

6 0
3 years ago
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