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3 years ago

Question 11 (1 point)

Engineering

1 answer:

kirill115 [55]3 years ago

4 0

Answer:

False

Explanation:

Bella counts products in finished goods inventory and she counts kits in various stages of manufacturing.

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3. What is the mechanical advantage of the pulley system shown below? HII TAI 190 O A1 O E.2 OC.3 OD 4

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Answer:

I don't know ☺️☺️☺️❌‼️

Explanation:

I don't understand this question

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2 years ago

Please help me with this. Plzzz.

Drupady [299]

Answer:

450,000m = 450km = 4.5E5

32,600,000W = 32.6MW = 3.26E7

59,700,000,000cal = 59.7Gcal = 5.97E10

0.000000083s = 83ns = 8.3E-8

35,000Ω = 35kΩ = 3.5E4

Explanation:

Giga = 1,000,000,000

Mega = 1,000,000

kilo = 1,000

unit = 1

deci = .1

centi = .01

milli = .001

micro = .000001

nano = .0000000001

pico = .000000000001

You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.

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3 years ago

How does warming up the tires on a car increase grip with the pavement?

ICE Princess25 [194]

Answer:

because burning rubber increases the grip power

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2 years ago

A flat plate 1.5 m long and 1.0 m wide is towed in water at 20 o C in the direction of its length at a speed of 15 cm/s. Determi

beks73 [17]

Answer:

15.8

0.0944

Explanation:

L = 1.5

B = 1.0

Speed of water = 15cm

Temperature = 20⁰C

At 20⁰C

Specific weight = 9790

Kinematic viscosity v = 1.00x10^-4m²/s

Dynamic viscosity u = 1.00x10^-3

Density p = 998kg/m²

Reynolds number

= 0.15x1.5/1.00x10^-4

= 225000

S = 5

5x1.5/225000^1/2

= 0.0158

= 15.8mm

Resistance on one side of plate

F = 0.664x1x1.0x10^-3x0.15x225000^1/2

= 0.04724N

Total resistance

= 2N

= 2x0.04724

= 0.0944N

3 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∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago