Answer:
The three ways thermal energy is transferred are;
1) Conduction
2) Convection
3) Radiation
Explanation:
1) The conduction of the heat from the open flame to the marshmallow is through the direct contact of the flame with the marshmallow, such that the flame the region of the combustion reaction, that produces light and heat touches the marshmallow
2) The convection process is the transfer of heat from the rising heated combustion products, as well as the heated air that rises from the flame
3) The radiation heat transfer is the transfer of the heat from the fire to the marshmallows directly by the heat the moves in the form of electromagnetic waves at temperatures above 1000 K, without the need for a medium, such that the marshmallow can be heated by the heat coming from side of the flame.
Answer:
The correct answer is B. Statements (i) and (ii) are true.
Explanation:
<u>Fisrt statement:</u>
Electromagnetic radiation such as wave, wavelength λ and oscillation frequency ν are related by a constant, the speed of light. The equation is given by:
So the first statement is true.
<u>Second statement:</u>
The value of c in the vacuum is 3×10⁸ m/s. Hence, the second statement is true.
<u>Third statement:</u>
The speed of any electromagnetic radiation is constant regardless the type of radiation.
Hence, the third statement is false.
Answer:
C = 771.35 J/kg°C
Explanation:
Here, e consider the conservation of energy equation. The conservation of energy principle states that:
Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container
Since,
Heat Given or Absorbed by a material = m C ΔT
Therefore,
m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃
where,
m₁ = Mass of Metal Piece = 2.3 kg
C = Specific Heat of Metal = ?
ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C
m₂ = Mass of Metal Container = 3.8 kg
ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C
m₃ = Mass of Water = 20 kg
C₃ = Specific Heat of Water = 4200 J/kg°C
ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C
Therefore,
(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)
C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J
C = 252000 J/326.7 kg°C
<u>C = 771.35 J/kg°C</u>
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
<h3>
Minimum force to be applied </h3>
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
- μ(s) is coefficient of static friction
- m is mass of the block
- g is acceleration due to gravity
F = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
<h3>Acceleration of the block</h3>
F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
Learn more about minimum force here: brainly.com/question/14353320
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Answer:
False.
Explanation:
The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,
FBA = −FAB
However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:
atruck = F/M
acar = F/m
and combining these we get:
atruck/acar = m/M
So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.
