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Ratling [72]
2 years ago
10

Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont

ainer filled with water, while the second one is allowed to cool naturally in the air. For which solid is the lumped system analysis more likely to be applicable? Why?
Physics
1 answer:
sergeinik [125]2 years ago
4 0
<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

  • If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable
  • The  reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air

Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body

∴ For a lumped system analysis Biot number should be less than 0.1

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if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^
Zarrin [17]
This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:

F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²]
4 0
2 years ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
2 years ago
A box of mass 3kg is lifted 1.5m onto a shelf. Calculate the change in its gravitational potential energy. The gravitational fie
Vedmedyk [2.9K]

Answer:

The change in gravitational potential energy is 45 J.

Explanation:

Given that,

Mass = 3 kg

Distance = 1.5 m

Gravitational field strength = 10 N/kg

We need to calculate the change in gravitational potential energy

Using formula of  gravitational potential energy

Change\ in\ gravitational\ potential\ energy =gravitational\ field\ strength\times mass\times distance

Put the value into the formula

Change\ in\ gravitational\ potential\ energy =10\times3\times1.5

Change\ in\ gravitational\ potential\ energy =45\ J

Hence, The change in gravitational potential energy is 45 J.

5 0
2 years ago
A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied
hammer [34]

Answer:

Explanation:

Reaction force of inclined surface = mg cosθ

= Friction force acting in upward direction = μ x mg cosθ

If F be force required in upward direction to keep the block at rest on the plane

F +  μ x mg cosθ = mg sinθ

F = mg sinθ -  μ x mg cosθ

F = mg( sinθ - μ cosθ)

= 2 x 9.8 ( sin34 - 0.2 cos34 )

= 19.6 ( .559 - .1658)

= 7.7 N

This is the minimum force required .

6 0
2 years ago
A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular
QveST [7]

Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J  

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J  

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J  

7 0
2 years ago
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