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Rzqust [24]
3 years ago
13

A wheel starts from rest at an angle of zero, and has

Physics
1 answer:
MrMuchimi3 years ago
4 0

Answer:

-22

Explanation:

Khan Academy

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Answer:

Explanation:

The direction of the acceleration is in the same direction as the net force causing it.  F = ma is actually a vector equation in which f and a are both vectors and m is a scalar constant.

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which of these chemical formulas represents two elements whose atoms have joined together with an ionic bond? A. O2 B. CaO C.NeN
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B: CaO
Because Calcium is a Metal and Oxygen is a Non-Metal. Ionic bonds are only formed with a Metal and a Non-Metal element.
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A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line
azamat

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
  • Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
  • Change in temperature (∆T) = 90 °F
  • Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

Learn more about coefficient of linear expansion:

brainly.com/question/28293570

#SPJ1

3 0
1 year ago
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
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