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How can you ensure safety while lifting heavy pots?
To lift heavy pots safely, seek assistance from others to distribute the weight. Use proper posture by bending your knees and keeping your back straight to prevent strain. Hold the pot firmly and ensure your grip is secure. Avoid overexertion and assess the weight before lifting. If unsure, ask forRead more
To lift heavy pots safely, seek assistance from others to distribute the weight. Use proper posture by bending your knees and keeping your back straight to prevent strain. Hold the pot firmly and ensure your grip is secure. Avoid overexertion and assess the weight before lifting. If unsure, ask for help or use equipment like trolleys to transport pots. These precautions prevent injuries and accidents during gardening activities.
See lessWhat are the steps to plant seeds in a kitchen garden?
Planting seeds in a kitchen garden involves preparing nutrient-rich soil and ensuring proper aeration. Make small holes or furrows for the seeds and plant them at the appropriate depth based on their type. Cover the seeds lightly with soil, water gently to moisten, and maintain consistent hydration.Read more
Planting seeds in a kitchen garden involves preparing nutrient-rich soil and ensuring proper aeration. Make small holes or furrows for the seeds and plant them at the appropriate depth based on their type. Cover the seeds lightly with soil, water gently to moisten, and maintain consistent hydration. Place the garden in a location with adequate sunlight, as it is essential for seed germination and healthy growth. Regular care ensures optimal results.
See lessWhy is manure or fertilizer important in gardening?
Manure and fertilizers play a crucial role in gardening by supplying vital nutrients, such as nitrogen, phosphorus, and potassium, which support healthy plant growth. They improve soil fertility, structure, and water retention, ensuring plants receive adequate nourishment. Organic manure enhances soRead more
Manure and fertilizers play a crucial role in gardening by supplying vital nutrients, such as nitrogen, phosphorus, and potassium, which support healthy plant growth. They improve soil fertility, structure, and water retention, ensuring plants receive adequate nourishment. Organic manure enhances soil health naturally, while fertilizers provide targeted nutrition for better yields. Regularly adding these to the garden ensures robust plant development and higher-quality produce, making them indispensable for successful gardening.
See lessHeat given to a body, which raises its temperature by 1° C is
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance. Definition: The specific heat c is mathematically defined as: c = Q / (m ΔT) Where: - c =Read more
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance.
Definition:
The specific heat c is mathematically defined as:
c = Q / (m ΔT)
Where:
– c = specific heat (J/kg°C or J/kg·K)
– Q = amount of heat added to the substance (in joules)
– m = mass of the substance (in kilograms)
– ΔT = change in temperature (in °C or K)
Explanation:
When heat Q is added to a substance, its temperature increases. The specific heat tells us how much heat is needed to achieve a specific temperature rise per unit mass of the substance.
For instance, if 500 joules of heat are added to 2 kg of water and the temperature increases by 10° C, then the specific heat can be calculated as follows:
1. Given:
Q = 500 J
m = 2 kg
ΔT = 10° C
2. Calculation of Specific Heat:
c = Q / (m ΔT) = 500 J / (2 kg × 10 °C) = 500 J / 20 kg°C = 25 J/kg°C
That is, 25 joules of heat raises the temperature of 1 kg of water by 1° C.
Conclusion:
Specific heat is an important concept in thermodynamics and material science because it helps determine how substances respond to heat transfer and temperature changes. It is crucial for applications such as heating, cooling, and thermal energy storage.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
A spherical black body with a radius 12 cm radiates 450W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as: P = σ A T⁴ Where: P is in watts and the Stefan-Boltzmann constant is approxiRead more
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as:
P = σ A T⁴
Where:
P is in watts and the Stefan-Boltzmann constant is approximately 5.67 × 10^(-8) W/m²K⁴ A = Surface area of the sphere
– T = absolute temperature (in Kelvin)
Given:
– Initial radius r₁ = 12 cm = 0.12 m
– Initial power P₁ = 450 W
– Initial temperature T₁ = 500 K
Surface Area Calculation:
The surface area A of a sphere is given by:
A = 4πr²
1. Initial Surface Area:
A₁ = 4π(0.12)² = 4π(0.0144) ≈ 0.18096 m²
2. New Conditions:
– The new radius r₂ = r₁/2 = 12/2 = 6 cm = 0.06 m
– The new temperature T₂ = 2 × T₁ = 2 × 500 = 1000 K
3. New Surface Area:
A₂ = 4π(0.06)² = 4π(0.0036) ≈ 0.04524 m²
New Power Calculation:
We can now determine the new power P₂ with the following expression:
P₂ = σ A₂ T₂⁴
Since the Stefan-Boltzmann constant σ is universal, we can write the new power in terms of the old power:
P₂/P₁ = (A₂ T₂⁴) / (A₁ T₁⁴)
We now substitute the values:
1. **Compute T₂⁴ and T₁⁴:
T₂⁴ = (1000)⁴ = 10¹²
T₁⁴ = (500)⁴ = 6.25 × 10¹¹
2. Compute the ratio:
P₂/P₁ = (0.04524 × 10¹²) / (0.18096 × 6.25 × 10¹¹)
Computing the above:
P₂/P₁ ≈ (0.04524 × 10¹²) / (1.12625 × 10¹¹) ≈ 0.401
3. Compute P₂:
Now, we put P₁ = 450 W:
P₂ = P₁ × (0.401) × (10¹² / 10¹¹) = 450 × 401 = 900 W
Conclusion:
Thus, if the radius were halved and the temperature doubled, the power radiated would be 900 W. The correct answer is 900.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/