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3 years ago

If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is

Engineering

2 answers:

xxMikexx [17]3 years ago

5 0

Answer:

90 degrees

Explanation:

In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane

Therefore as per the given situation it should be 90 degrees from the plane

hence, the same is to be considered and relevant too

serious [3.7K]3 years ago

5 0

90 becaus I’m the goat

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Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and

ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, $Q=16 m^{3}/min$

Diameter d= 8cm= 0.08 m

Assumptions

The flow is jet flow hence momentum-flux correction factor is unity
Gravitational force is not considered
The flow is steady, frictionless and incompressible
Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

$V=\frac {Q}{A}$ where V is the exit velocity and A is area

Area, $A=\frac {\pi d^{2}{4}$ where d is the diameter

By substitution

$V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min$

To convert v to m/s from m/s, we simply divide it by 60 hence

$V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s$

3 0

3 years ago

A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti

natima [27]

Answer:

Final mass of Argon= 2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

= (200 * 303 ) / (450 * 219 ) * 4

= 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0

3 years ago

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value

OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^{-kt^n}$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^{-k^146^2.1}$

$0.50 = e^{-k306.6}$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26\times 10^{-3}$

for 86 % completion

$0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}$

$e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14$

$-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)$

$t^{2.1} = 869.96$

t = 25.10 sec

5 0

3 years ago

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

6 0

3 years ago

How does a truth tables impact the development of a completed circuit? Make sure to use as many keywords as possible to back up

WINSTONCH [101]

Answer:

it is one this so the key word are pfp tpf htf pfp

Explanation:

3 0

3 years ago