The first step in treating shock is to

A flow of 12 m/s passes through a 6 m wide, 2 m deep rectangular channel with a bed slope of 0. 001. If the mean velocity of flo

prohojiy [21]

Answer:

manning's coefficient is 0.0357

Explanation:

Given:

Velocity of flow, v = 12 m/s

Width of the channel, b = 6 m

Depth of the channel, d = 2 m

bed slope, s = 0.001

mean velocity of flow, V = 1 m/s

now, the velocity is given as:

$V= \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}}$

where,

n is the manning's coefficient

R is hydraulic mean depth

R = (Area of the channel) / (wetted Perimeter of the channel)

now,

R = (2 × 6) / ((2 × 2) + 6)

or

R = 12 / 10 = 1.2 m

now, on substituting the values in the equation for velocity, we get

$1= \frac{1}{n}1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}$

or

$n= 1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}$

or

n = 0.0357

hence, the value of manning's coefficient is 0.0357

8 0

3 years ago

At the instant shown car A is travelling with a velocity of 24 m/s and which is decreasing at 4 m/s2 along the highway. At the s

SVEN [57.7K]

(a) V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

<u>Explanation:</u>

We will solve with respect to Cartesian vector form.

So,

V(A)= (24i) m/s

acc(A) = (4i) m/s²

There are two components of Car B, cos 60⁰ and sin 60⁰

V(B) = 20 cos 60° i + 20 sin 60° j

V(B) = (10 i + 17.32 j ) m/s

The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.

The tangential acceleration, a(t) = 5 m/s²

Normal acceleration, a(n) = $\frac{v^2}{p} \\\\$

So,

$a(n) = \frac{(20)^2}{250}\\ \\a(n) = 1.6 m/s^2$

For the tangential acceleration, the acceleration is slowing down. So,

a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²

a(t) = ( -2.5 i - 4.33 j) m/s²

For normal acceleration, it towards center. So,

a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²

a(n) = (1.39 i - 0.8 j ) m/s²

Total acceleration of Car B:

acc(B) = a(t) + a(n)

acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²

acc(B) = (-1.11i - 5.13 j ) m/s²

(a) V(A/B) = ?

V(A) = V(B) + V(A/B)

(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)

V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ?

acc(A) = acc(B) + acc(A/B)

(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)

acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

3 0

3 years ago

The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes

kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0

4 years ago

3. If nothing can ever be at absolute zero, why does the concept exist?

Tanzania [10]

The absolute zero in temperature refers to the minimal possible temperature. It is the temperature at which the molecules of a system stop moving, so it is a really useful reference point.

<h3>Why absolute zero can't be reached?</h3>

It would mean that we need to remove all the energy from a system, but to do this we need to interact with the system in some way, and by interacting with it we give it "some" energy.

Actually, from a quantum mechanical point of view, the absolute zero has a residual energy (so it is not actually zero) and it is called the "zero point". This happens because it must meet <u>Heisenberg's uncertainty principle</u>.

So yes, the absolute zero can't be reached, but there are really good approximations (At the moment there is a difference of about 150 nanokelvins between the absolute zero and the smallest temperature reached). Also, there are a lot of investigations near the absolute zero, like people that try to reach it or people that just need to work with really low temperatures, like in type I superconductors.

So, concluding, why does the concept exist?

Because it is a reference point.
It is the theoretical temperature at which the molecules stop moving, defining this as the <u>minimum possible temperature.</u>

If you want to learn more about the absolute zero, you can read:

brainly.com/question/3795971

3 0

3 years ago

An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit

kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²) ......................1

here ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99 .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

put here value

0.005 = 120/k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

k ×∈ × √(1-2∈²) = 1200 ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0

3 years ago