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8_murik_8 [283]
3 years ago
5

Heat always flows from a _________ object to a __________ one.

Physics
2 answers:
PolarNik [594]3 years ago
8 0

From a hot object to a cold object. Ill give an example as to why:

Say you place an ice cube in a glass of warm water, we know that in colder substances, the average kinetic energy per particle (which is the classical definition for temperature is lower). In hotter substances the average kinetic energy per particle is larger, as it is hotter. So now the ice is placed in the warm water. The fast moving particles within the water, that have higher kinetic energies will bounce off the slower moving particles in the ice with lower average kinetic energies. This causes an exchange in energy as the substance with more average kinetic energy per particle (which is temperature) will transfer energy to the slower moving particles located in the ice. This will eventually cause the water and ice to reach an equilibrium temperature as a result of the exchange, with the energy taken to melt the ice included.

satela [25.4K]3 years ago
6 0

Heat always flows from a B. hotter object to a colder one.

Heat transfers always from hotter to colder.

~

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Which has the same meaning as air pressure?
SVETLANKA909090 [29]
B) atmosphere pressure, i believe

7 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
An object moves from the position +16m to the position +43m in 12s. What us the total displacement
NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

                                     =27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0
3 years ago
Why is a protective apron or lab coat important to use when working with acids?
Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

  • An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
  • A strong acid ionizes completely in solution.
  • When they come in contact with a fabric, they break them down violently.
  • So, if they come in contact with the skin, it causes a violent break down of body tissues.
  • The apron acts a protective layer.
4 0
3 years ago
Read 2 more answers
. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt
Vinil7 [7]

Answer:

a)    L = 3.29 10⁻⁴ H,  b)U = 5.33 10⁻²  J

Explanation:

a) The inductance is a solenoid this given carrier

           L = \frac{N \ \phi_B }{I}

The magnetic field inside the solenoid is

          B = μ₀ \frac{N}{l}  i

hence the magnetic flux

          Ф_B = B. A = μ₀ \frac{N \ A}{l \ i}

we substitute in the expression of inductance

          L = N² μ₀ A /l

let's find the area of ​​each turn

          A = π r²

         A = π 0.02²

         A = 1.2566 10⁻³ m²

let's calculate

          L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

          L = 3.29 10⁻⁴ H

b) The stored energy is

           U = ½ L i²

let's calculate

            U = ½ 3.29 10⁻⁴ 18²

            U = 5.33 10⁻²  J

5 0
3 years ago
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