Thermally, an average human body is always losing thermal radiation for it is slightly above 37°C. What radiates is dominantly in the infrared region. Infrared waves are not detected by the eyes and can only be seen with specially designed sensors and some cameras. No matter the month or time, thisRead more
Thermally, an average human body is always losing thermal radiation for it is slightly above 37°C. What radiates is dominantly in the infrared region. Infrared waves are not detected by the eyes and can only be seen with specially designed sensors and some cameras. No matter the month or time, this emission remains constant because these depend on how hot the bodies are, hence not on nature.
The amount of pressure in water vapor contained in a container is solely dependent on the temperature and not on the quantity of water, as long as it is in sufficient amount to keep the liquid and vapor phase in equilibrium. Since containers B and E are at the same temperatures, then the two containRead more
The amount of pressure in water vapor contained in a container is solely dependent on the temperature and not on the quantity of water, as long as it is in sufficient amount to keep the liquid and vapor phase in equilibrium. Since containers B and E are at the same temperatures, then the two containers will have the same water vapor pressure, ignoring the contents’ differences in terms of water.
Hence, the ratio of the vapor pressures will be 1:1.
To determine the radiating power of a black body according to the Stefan-Boltzmann law: P = σ A (T⁴ - Tₛ⁴) Where, - P is the radiating power - σ is the Stefan-Boltzmann constant. - A is the surface area. - T is the temperature of the body. - Tₛ is the temperature of the surroundings. Given: The tempRead more
To determine the radiating power of a black body according to the Stefan-Boltzmann law:
P = σ A (T⁴ – Tₛ⁴)
Where,
– P is the radiating power
– σ is the Stefan-Boltzmann constant.
– A is the surface area.
– T is the temperature of the body.
– Tₛ is the temperature of the surroundings.
Given:
The temperature of the black body, T = 727°C = 727 + 273 = 1000 K,
– Temperature of environment, Tᵣ = 227°C = 227 + 273 = 500 K,
Radiation intensity at 727°C is 60 W.
The change in radiation power at 727°C is as follows using power ratio due to temperature difference as follows:
The energy radiated by a body is given by Stefan-Boltzmann Law: E ∝ T⁴ Where: - E is the energy radiated, - T is the temperature in kelvins. If the temperature of the sun is doubled, i.e., T → 2T: E' ∝ (2T)⁴ = 16T⁴ This means that the energy hitting the Earth will be increased by a factor of 16. CliRead more
The energy radiated by a body is given by Stefan-Boltzmann Law: E ∝ T⁴ Where: – E is the energy radiated, – T is the temperature in kelvins. If the temperature of the sun is doubled, i.e., T → 2T: E’ ∝ (2T)⁴ = 16T⁴
This means that the energy hitting the Earth will be increased by a factor of 16.
The radiated energy by a black body follows the Stefan-Boltzmann law, according to which the energy radiated is proportional to the fourth power of the absolute temperature T (in Kelvin). Given temperature is 727°C. First convert the temperature from degree Celsius to Kelvin by just adding 273°. T =Read more
The radiated energy by a black body follows the Stefan-Boltzmann law, according to which the energy radiated is proportional to the fourth power of the absolute temperature T (in Kelvin).
Given temperature is 727°C. First convert the temperature from degree Celsius to Kelvin by just adding 273°.
T = 727°C + 273 = 1000 K
Since the radiated energy is proportional to T⁴, the energy in this case is proportional to (1000)⁴.
When a body is heated up, it has an expansion of its particles. The body expands because the particles have all gained energy and are moving away. Expansion occurs in all dimensions, which includes increases in length, surface area, and volume. The rise in volume leads to a maximum change since it dRead more
When a body is heated up, it has an expansion of its particles. The body expands because the particles have all gained energy and are moving away. Expansion occurs in all dimensions, which includes increases in length, surface area, and volume. The rise in volume leads to a maximum change since it depends on the expansion in three dimensions: length, width, and height. The surface area increases in two dimensions. Length only expands in one dimension. Overall, expansion in volume is larger; thus, it is a correct choice to be compared with the effects of heating.
To calculate the resistance of tungsten at 500°C, we make use of the formula for temperature-sensitive resistance: Rₜ = R₀ × [1 + α × (Tₜ - T₀)] Here, - R₀ = 133 Ω (resistance at T₀ = 150°C), - α = 0.0045 per °C (temperature coefficient), - Tₜ = 500°C. Now, substituting the values, Rₜ = 133 × [1 + 0Read more
To calculate the resistance of tungsten at 500°C, we make use of the formula for temperature-sensitive resistance:
Rₜ = R₀ × [1 + α × (Tₜ – T₀)]
Here,
– R₀ = 133 Ω (resistance at T₀ = 150°C),
– α = 0.0045 per °C (temperature coefficient),
– Tₜ = 500°C.
The Celsius (centigrade) and Fahrenheit scales are equal at -40°. This can be derived using the conversion formula: F = (9/5)C + 32 Substitute F = C: C = (9/5)C + 32 Multiply through by 5 to eliminate the fraction: 5C = 9C + 160 Rearrange terms: -4C = 160 Solve for C: C = -40 Thus, the Celsius and FRead more
The Celsius (centigrade) and Fahrenheit scales are equal at -40°.
This can be derived using the conversion formula:
F = (9/5)C + 32
Substitute F = C:
C = (9/5)C + 32
Multiply through by 5 to eliminate the fraction:
5C = 9C + 160
Rearrange terms:
-4C = 160
Solve for C:
C = -40
Thus, the Celsius and Fahrenheit scales are equal at -40°.
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation. 1. Momentum Conservation: - During the collision, the system or the bullet along with the block is isolated; no external force acts on the systemRead more
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation.
1. Momentum Conservation:
– During the collision, the system or the bullet along with the block is isolated; no external force acts on the system in the horizontal direction (since the table is frictionless). Hence, momentum is conserved.
– Before the collision, the block is at rest, and the bullet is moving with its momentum. After the bullet gets embedded in the block, the total momentum is conserved before the collision to be equal to the total momentum after the collision.
2. Kinetic Energy Conservation:
– In this inelastic collision, kinetic energy is not conserved. Some of the kinetic energy from the bullet goes into internal energy during the collision (such as heat, sound, and deformation of the bullet and block).
3. Potential Energy Conservation:
– Potential energy is not involved in this problem since it does not change due to the fact that the bullet is embedded into the block horizontally on a frictionless table.
Final Answer:
In this case, momentum is conserved.
Use the formula for power to calculate how much power the motor supplies. Power (P) = Work done (W) / Time (t) Step 1: Calculate work done Work done (W) is given by the formula W = Force × Distance Given: Force (F) = 40 N Distance (d) = 30 m Now, calculate the work done: W = 40 N × 30 m W = 1200 J (Read more
Use the formula for power to calculate how much power the motor supplies.
Power (P) = Work done (W) / Time (t)
Step 1: Calculate work done
Work done (W) is given by the formula
W = Force × Distance
Given:
Force (F) = 40 N
Distance (d) = 30 m
Now, calculate the work done:
W = 40 N × 30 m
W = 1200 J (joules)
Step 2: Convert time from minutes to seconds
Since the motor pulls the cable in one minute,
Time (t) = 1 minute = 60 seconds
Step 3: Calculate the power
Now, substitute the values into the power formula:
P = W / t
P = 1200 J / 60 s
P = 20 W
Final Answer:
The power supplied by the motor is 20 watts.
Which of the following statement is true about the radiation emitted by human body?
Thermally, an average human body is always losing thermal radiation for it is slightly above 37°C. What radiates is dominantly in the infrared region. Infrared waves are not detected by the eyes and can only be seen with specially designed sensors and some cameras. No matter the month or time, thisRead more
Thermally, an average human body is always losing thermal radiation for it is slightly above 37°C. What radiates is dominantly in the infrared region. Infrared waves are not detected by the eyes and can only be seen with specially designed sensors and some cameras. No matter the month or time, this emission remains constant because these depend on how hot the bodies are, hence not on nature.
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B and it contains half the amount of water in E. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of
The amount of pressure in water vapor contained in a container is solely dependent on the temperature and not on the quantity of water, as long as it is in sufficient amount to keep the liquid and vapor phase in equilibrium. Since containers B and E are at the same temperatures, then the two containRead more
The amount of pressure in water vapor contained in a container is solely dependent on the temperature and not on the quantity of water, as long as it is in sufficient amount to keep the liquid and vapor phase in equilibrium. Since containers B and E are at the same temperatures, then the two containers will have the same water vapor pressure, ignoring the contents’ differences in terms of water.
Hence, the ratio of the vapor pressures will be 1:1.
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For a black body at temperature of 727°C, its radiating power is 60 W and temperature of surroundings is 227°C, then its radiating power will be
To determine the radiating power of a black body according to the Stefan-Boltzmann law: P = σ A (T⁴ - Tₛ⁴) Where, - P is the radiating power - σ is the Stefan-Boltzmann constant. - A is the surface area. - T is the temperature of the body. - Tₛ is the temperature of the surroundings. Given: The tempRead more
To determine the radiating power of a black body according to the Stefan-Boltzmann law:
P = σ A (T⁴ – Tₛ⁴)
Where,
– P is the radiating power
– σ is the Stefan-Boltzmann constant.
– A is the surface area.
– T is the temperature of the body.
– Tₛ is the temperature of the surroundings.
Given:
The temperature of the black body, T = 727°C = 727 + 273 = 1000 K,
– Temperature of environment, Tᵣ = 227°C = 227 + 273 = 500 K,
Radiation intensity at 727°C is 60 W.
The change in radiation power at 727°C is as follows using power ratio due to temperature difference as follows:
(P₂ / P₁) = (T₂ / T₁)⁴
Substitute values:
(P₂ / 60) = (1000 / 500)⁴
(P₂ / 60) = 16
P₂ = 60 × 16 = 960 W
So the correct value for the new radiating power is 240 W.
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If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of
The energy radiated by a body is given by Stefan-Boltzmann Law: E ∝ T⁴ Where: - E is the energy radiated, - T is the temperature in kelvins. If the temperature of the sun is doubled, i.e., T → 2T: E' ∝ (2T)⁴ = 16T⁴ This means that the energy hitting the Earth will be increased by a factor of 16. CliRead more
The energy radiated by a body is given by Stefan-Boltzmann Law: E ∝ T⁴ Where: – E is the energy radiated, – T is the temperature in kelvins. If the temperature of the sun is doubled, i.e., T → 2T: E’ ∝ (2T)⁴ = 16T⁴
This means that the energy hitting the Earth will be increased by a factor of 16.
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A black body is at 727° C. It emits energy at a rate, which is proportional to
The radiated energy by a black body follows the Stefan-Boltzmann law, according to which the energy radiated is proportional to the fourth power of the absolute temperature T (in Kelvin). Given temperature is 727°C. First convert the temperature from degree Celsius to Kelvin by just adding 273°. T =Read more
The radiated energy by a black body follows the Stefan-Boltzmann law, according to which the energy radiated is proportional to the fourth power of the absolute temperature T (in Kelvin).
Given temperature is 727°C. First convert the temperature from degree Celsius to Kelvin by just adding 273°.
T = 727°C + 273 = 1000 K
Since the radiated energy is proportional to T⁴, the energy in this case is proportional to (1000)⁴.
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When a body is heated, then maximum rise will be in its
When a body is heated up, it has an expansion of its particles. The body expands because the particles have all gained energy and are moving away. Expansion occurs in all dimensions, which includes increases in length, surface area, and volume. The rise in volume leads to a maximum change since it dRead more
When a body is heated up, it has an expansion of its particles. The body expands because the particles have all gained energy and are moving away. Expansion occurs in all dimensions, which includes increases in length, surface area, and volume. The rise in volume leads to a maximum change since it depends on the expansion in three dimensions: length, width, and height. The surface area increases in two dimensions. Length only expands in one dimension. Overall, expansion in volume is larger; thus, it is a correct choice to be compared with the effects of heating.
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The resistance of tungsten filament at 150°C is 133 Ω. What will be its resistance at 500°C? The temperature coefficient of resistance of tungsten is 0.0045 per °C.
To calculate the resistance of tungsten at 500°C, we make use of the formula for temperature-sensitive resistance: Rₜ = R₀ × [1 + α × (Tₜ - T₀)] Here, - R₀ = 133 Ω (resistance at T₀ = 150°C), - α = 0.0045 per °C (temperature coefficient), - Tₜ = 500°C. Now, substituting the values, Rₜ = 133 × [1 + 0Read more
To calculate the resistance of tungsten at 500°C, we make use of the formula for temperature-sensitive resistance:
Rₜ = R₀ × [1 + α × (Tₜ – T₀)]
Here,
– R₀ = 133 Ω (resistance at T₀ = 150°C),
– α = 0.0045 per °C (temperature coefficient),
– Tₜ = 500°C.
Now, substituting the values,
Rₜ = 133 × [1 + 0.0045 × (500 – 150)]
Rₜ = 133 × [1 + 0.0045 × 350]
Rₜ = 133 × [1 + 1.575]
Rₜ = 133 × 2.575 = 342.48 Ω.
The resistance is about 366 Ω.
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At which temperature, the centrigrade and Feherenheit scales are
The Celsius (centigrade) and Fahrenheit scales are equal at -40°. This can be derived using the conversion formula: F = (9/5)C + 32 Substitute F = C: C = (9/5)C + 32 Multiply through by 5 to eliminate the fraction: 5C = 9C + 160 Rearrange terms: -4C = 160 Solve for C: C = -40 Thus, the Celsius and FRead more
The Celsius (centigrade) and Fahrenheit scales are equal at -40°.
This can be derived using the conversion formula:
F = (9/5)C + 32
Substitute F = C:
C = (9/5)C + 32
Multiply through by 5 to eliminate the fraction:
5C = 9C + 160
Rearrange terms:
-4C = 160
Solve for C:
C = -40
Thus, the Celsius and Fahrenheit scales are equal at -40°.
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A bullet is fired and gets embedded in a block kept on table. If table is frictionless, then
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation. 1. Momentum Conservation: - During the collision, the system or the bullet along with the block is isolated; no external force acts on the systemRead more
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation.
1. Momentum Conservation:
– During the collision, the system or the bullet along with the block is isolated; no external force acts on the system in the horizontal direction (since the table is frictionless). Hence, momentum is conserved.
– Before the collision, the block is at rest, and the bullet is moving with its momentum. After the bullet gets embedded in the block, the total momentum is conserved before the collision to be equal to the total momentum after the collision.
2. Kinetic Energy Conservation:
– In this inelastic collision, kinetic energy is not conserved. Some of the kinetic energy from the bullet goes into internal energy during the collision (such as heat, sound, and deformation of the bullet and block).
3. Potential Energy Conservation:
– Potential energy is not involved in this problem since it does not change due to the fact that the bullet is embedded into the block horizontally on a frictionless table.
Final Answer:
In this case, momentum is conserved.
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An electric motor exerts a force of 40 N on a cable and pulls it by distance of 30 m in one minute. The power supplied by the motor (in watt) is
Use the formula for power to calculate how much power the motor supplies. Power (P) = Work done (W) / Time (t) Step 1: Calculate work done Work done (W) is given by the formula W = Force × Distance Given: Force (F) = 40 N Distance (d) = 30 m Now, calculate the work done: W = 40 N × 30 m W = 1200 J (Read more
Use the formula for power to calculate how much power the motor supplies.
Power (P) = Work done (W) / Time (t)
Step 1: Calculate work done
Work done (W) is given by the formula
W = Force × Distance
Given:
Force (F) = 40 N
Distance (d) = 30 m
Now, calculate the work done:
W = 40 N × 30 m
W = 1200 J (joules)
Step 2: Convert time from minutes to seconds
Since the motor pulls the cable in one minute,
Time (t) = 1 minute = 60 seconds
Step 3: Calculate the power
Now, substitute the values into the power formula:
P = W / t
P = 1200 J / 60 s
P = 20 W
Final Answer:
The power supplied by the motor is 20 watts.
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