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NikAS [45]
4 years ago
11

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

aced within a channel to maintain an upstream depth of 2.5 m for a flow of 0.5 m3/s
Physics
1 answer:
Alex4 years ago
7 0

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

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What do you mean by average velocity​
mina [271]

Answer:

Here is the answer. Hope this helps you!

Explanation:

Average velocity is the sum of initial and final velocity divided by 2. I t is the same as total Displacement divided by total time. Average velocity is calculated when the body is in non-uniform motion (also when total displacement and time is not given). The formula is as follows:

u + v/2 = v_{av}

Average velocity = Average speed

during motion in a straight line. therefore, the above mentioned formula can be used for calculating average speed as well, when the direction is one and only the same, that is, during motion in a straight line. The S.I unit remains the same-

m/s.

Since velocity is vector, average VELOCITY is also vector. However, Average SPEED is scalar as speed is scalar. both can be equal only when the distance = displacement and when they are following the same direction of motion.

7 0
3 years ago
A charge of 5.4 C experiences a force of 25.0 in an electric field. What is the strength of electric field at that point ? If th
Airida [17]

The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

<h3></h3><h3>What is the electric field strength?</h3>

The electric field strength is defined as the ratio of electric force to charge.

Given data;

q₁ = 5.4 C

F₁ is the electric force in case1

E is the electric field =?

F₂ is the electric force in case 2

q₂ is the charge 2

The strength of the electric field at that point is;

F₁=Eq₁

E₁=F/q₁

E₁=25.0 N / 5.4 C

E₁=4.587 N/C

The force would this charge experience at that point when the charge is 2.7 C;

F₂=Eq₂

F₂=4.587 N/C × 2.7 C

F₂ = 12.38 N

Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

#SPJ1

8 0
2 years ago
Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat
fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

<h3 /><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.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625  N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ4

6 0
2 years ago
Figure 1 shows cavity wall insulation being installed in the wall of a house.
Alexeev081 [22]

Explanation:

Energy loss through walls can be reduced using cavity wall insulation. This involves blowing insulating material into the gap between the brick and the inside wall, which reduces the loss by air circulating inside the cavity, therefore reducing loss by convection through the cavity.

8 0
2 years ago
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

3 0
3 years ago
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