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Mademuasel [1]
3 years ago
5

Distance between Ranjan and Gomes house is 9km. Ranjan has to attend Gomes birthday party at 7 o'clock. He started from his home

at 6 o'clock on his bicycle and covered a distance of 6km in 40 minutes. At that point he met David and he spoke to him for 5minutes and reached Gomes birthday party at 7 o'clock. At what speed did he cover the second part of the journey? Calculate his average speed for the entire journey.Required to answer.Single line text.
Physics
1 answer:
Nataliya [291]3 years ago
5 0

Answer:

Kindly check explanation

Explanation:

Given the following :

Distance between Ranjan and Gomez house = 9km

Start time = 6'o clock

Time he arrived at Gomes house = 7'o clock

Time used for chatting during this period = 5 minutes

Distance covered in 40minutes = 6km

Time spend for second part of the journey = (60 - (40 + 5))minutes = 15 minutes = 15/60 = 0.25 hoir

Distance covered during second part of journey = 9 - 6 = 3 km

Speed = distance / time

Speed = 3km / 0.25 hr = 12km/hr

Average speed for entire journey :

First part :

Speed = distance / time

Time = 40 minutes = (40/60) = 0.667 hour

Speed = 6km / 0.667 hour = 9km / hr

Average speed :

(First part + second part) / 2

(9km/hr + 12km/hr) / 2 = 21km/hr / 2 = 10.5 km/hr

= 10.5km/hr

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

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

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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

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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).

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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.

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3 years ago
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a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

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