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2 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Engineering

1 answer:

Eva8 [605]2 years ago

6 0

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of Pressure is $M^{1}L^{-1}T^{-2}$
Dimensional formula of length is L
Dimensional formula of mass is M
Dimensional formula of velocity is $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

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All of the dimensions on an aircraft drawing are_________ to the bottom of the drawing.

Jet001 [13]

All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing

Answer: parallel

7 0

2 years ago

The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The b

Wewaii [24]

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁ and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0 (+↑)

Nef*Cos 12º - F₂*Sin 12º = W

⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18

⇒ Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒ P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

7 0

3 years ago

Read 2 more answers

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

Paraphin [41]

Answer:

$10.8\ \text{lb/ft^2}$

$101.96\ \text{lb/ft}^2$

Explanation:

$v_1$ = Velocity of car = 65 mph = $65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}$

$\rho$ = Density of air = $0.00237\ \text{slug/ft}^3$

$v_2=0$

$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

$P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}$

The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

$P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2$

The maximum pressure on the girl's hand is $101.96\ \text{lb/ft}^2$

5 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$