3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13

svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

Steady operations exist;
The heat transfer coefficient (h) is uniform over the entire fin surfaces;
Thermal conductivity (k) is constant;
Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0

3 years ago

Which of the following is an example of a chemical property? Density Solubility Flammability Magnetism

choli [55]

The answer should be flammability

8 0

4 years ago

7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

6 0

4 years ago

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

4 years ago

What is the difference between brake horsepower and thrust horsepower?

arlik [135]

1. HP is the output horsepower rating of an engine, while Brake horse power is the input brake horsepower of an engine. ... Brake horse power is the measurement of an engine's power without any power losses, while HP is less the power losses Brake horse power

5 0

3 years ago