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Shkiper50 [21]
3 years ago
13

1. The nearest distance of distinct vision of

Physics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

The focal length of the lens is +30 and it is a convex lens.

Explanation:

Given that,

The nearest distance of distinct vision of  a hypermetropial person is 60 cm, v = -60 cm

The distance is reduced by 20 cm, u = -20 cm

We need to find the nature and focal length of the lens.

Let f be the focal length of the lens. Using lens formula,

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{f}=\dfrac{1}{(-60)}-\dfrac{1}{(-20)}\\\\f=+30\ cm

So, the focal length of the lens is +30 and it is a convex lens.

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A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
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Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

time taken by the wave to reach the top is given as

t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

t = 12 s

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A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning
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Answer:

The tiger would not be able to produce glucose causing it to stop running

Explanation:

Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.

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(06.04 MC)<br> Which of the following best describes a human egg cell? (3 points)
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A car drives 20 miles north and then 3 miles south. what is the displacement or the car
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Read 2 more answers
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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