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julia-pushkina [17]
3 years ago
9

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

the magnitude
Physics
1 answer:
AveGali [126]3 years ago
3 0

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

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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

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.

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