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Dennis_Churaev [7]
2 years ago
8

HELP ASAP VERY IMPORTANT.

Physics
1 answer:
sertanlavr [38]2 years ago
4 0

<em>Answers:</em>

<em>1. The hot soup will loose the heat and the ice water will gain the heat</em>

  • If two jars are insulated inside the insulated box, the heat may not be transferred to outside of the box.
  • According to II law of thermodynamics, Heat always flow from high temperature body to low temperature body, with out aid of external energy.

<em>So, from two points, it is concluded that The hot soup will lose heat and the ice water will gain the heat until they reach the thermodynamic equilibrium.</em>

<em>2. The particles in gases are farther apart and move faster   </em>

  • Particles in the gases are loosely packed (greater distance between particles compared to solids and liquids) and particles collide less often.

<em>Therefore conduction is weak in gases compared to solids and liquids.</em>

<em>3. Heat and milk by conduction; popping popcorn by radiation.</em>

  • The heat can transfer from pot to the milk by conduction because they are in contact at boundaries, similarly the pot and the stove are in contact <em>so the conduction transfers heat from pot to the milk. </em>
  • In microwave oven there is no direct contact (<em>no conduction</em>) of heat and popcorn, also there is no molecular momentum transfer <em>(means of no convection).</em>

<em>So obviously the heat transfer by radiation occurs in a microwave oven.</em>

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Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t
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Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

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B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

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