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DerKrebs [107]
3 years ago
10

Calculating Potential Energy

Engineering
1 answer:
Ahat [919]3 years ago
7 0

Answer:

D. 539,000 J

Explanation:

The formula for potential energy(P.E) =mgh where m is mass of object, g is gravity acceleration 9.8 m/s² and h is height in meters.

From the question ;

m=500 kg

g= 9.8 m/s²

h= 110 m

P.E = mgh

     = 500*9.8*110

     =539,000 J

 Answer option D

The other options are incorrect because the calculated values for the potential energy of the train are not equal to 539,000 J.

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You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
Two vehicles arrive at an uncontrolled intersection from different streets at the same time 1.The driver on the right must yield
ss7ja [257]
3. Both vehicles must stop
4 0
3 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
35 points an brainiest if correct
Jet001 [13]
I think the answer is B. 10D
6 0
3 years ago
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