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Black holes in orbit around a normal star are detected gas into the black hole, which can reach temperatures greater than 10⁶K. Assuming that the infalling gas can be modelled as a black body radiator, then the wavelength of maximum power lies
We use Wien's Displacement Law to find the wavelength of maximum power emission for a black body: ln λₘₐₓ = b / T where b is Wien's constant, approximately 2.9 × 10⁻³ m·K, and T is the temperature in Kelvin. We know that the gas temperature is T > 10⁶ K, so we compute: λₘₐₓ = (2.9 × 10⁻⁹ m) / (10Read more
We use Wien’s Displacement Law to find the wavelength of maximum power emission for a black body: ln
λₘₐₓ = b / T
where b is Wien’s constant, approximately 2.9 × 10⁻³ m·K, and T is the temperature in Kelvin. We know that the gas temperature is T > 10⁶ K, so we compute:
λₘₐₓ = (2.9 × 10⁻⁹ m) / (10⁶ K)
λₘₐₓ = 2.9 nm
Since X-rays have wavelengths in the range of 0.01 nm to 10 nm, the maximum emission falls in the X-ray region.
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The area of the region bounded by the curve y = √x -1 and the lines x = 1 and x = 5 is
To determine the area of the region bounded by the curve y = √x - 1 and the lines x = 1 and x = 5, we must set up the definite integral. The area is given by: A = ∫₁⁵ (√x - 1) dx Step 1: Solve the integral We can split the integral into two parts: A = ∫₁⁵ √x dx - ∫₁⁵ 1 dx First integral: ∫ √x dx = ∫Read more
To determine the area of the region bounded by the curve y = √x – 1 and the lines x = 1 and x = 5, we must set up the definite integral.
The area is given by:
A = ∫₁⁵ (√x – 1) dx
Step 1: Solve the integral
We can split the integral into two parts:
A = ∫₁⁵ √x dx – ∫₁⁵ 1 dx
First integral:
∫ √x dx = ∫ x^(1/2) dx = (2/3) x^(3/2)
Evaluating this from 1 to 5:
[(2/3) x^(3/2)]₁⁵ = (2/3) (5^(3/2) – 1^(3/2)) = (2/3) (5√5 – 1)
Second integral:
∫ 1 dx = x
Evaluating this from 1 to 5:
[x]₁⁵ = 5 – 1 = 4
Step 2: Combine the results
The total area is:
A = (2/3) (5√5 – 1) – 4
Simplifying this expression gives the final result:
A = 13/3 square units
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The area of the region bounded by the curve x = 2y + 3 and the lines y = 1, y = -1 and y-axis is
In order to discover the area of the region bounded by the curve x = 2y + 3, the lines y = 1, y = -1, and the y-axis, we should set up the definite integral. Step 1: Determine the x-intercepts of the curve The curve is defined by the equation x = 2y + 3. For the y-axis, we let x = 0: 0 = 2y + 3 SolvRead more
In order to discover the area of the region bounded by the curve x = 2y + 3, the lines y = 1, y = -1, and the y-axis, we should set up the definite integral.
Step 1: Determine the x-intercepts of the curve
The curve is defined by the equation x = 2y + 3. For the y-axis, we let x = 0:
0 = 2y + 3
Solve for y:
So the curve crosses the y-axis at y = -3/2.
Step 2: Put the integral
The area is bounded by y = 1, y = -1, and the curve. We’ll have to integrate the function w.r.t to y in the range from y = -1 to y = 1 for finding the area,
A = ∫₋₁¹ (2y + 3) dy
Step 3: Evaluate the integral
Integrate the expression 2y + 3 first.
∫ (2y + 3) dy = y² + 3y
Now, substitute these values into this from y = -1 to y = 1:
A = [y² + 3y]₋₁¹
When y = 1:
1² + 3(1) = 1 + 3 = 4
When y = -1:
(-1)² + 3(-1) = 1 – 3 = -2
Thus the area is,
A = 4 – (-2) = 4 + 2 = 6
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The area of the region bounded by the curve y = x + 1 and the lines x = 2, x = 3 and x-axis in sq. units is
In order to find the area of the region determined by the curve y = x + 1 along with the lines x = 2, x = 3 and the x-axis, we first need to set up the definite integral. Step 1: Set up the integral To calculate the area, we will integrate the function y = x + 1 with respect to x between the limitsRead more
In order to find the area of the region determined by the curve y = x + 1 along with the lines x = 2, x = 3 and the x-axis, we first need to set up the definite integral.
Step 1: Set up the integral
To calculate the area, we will integrate the function y = x + 1 with respect to x between the limits x = 2 and x = 3.
A = ∫₂³ (x + 1) dx
Step 2: Integrate the function
First, integrate (x + 1):
∫ (x + 1) dx = (x²)/2 + x
Now, evaluate this from x = 2 to x = 3:
At x = 3:
(3²)/2 + 3 = 9/2 + 3 = 9/2 + 6/2 = 15/2
At x = 2:
(2²)/2 + 2 = 4/2 + 2 = 2 + 2 = 4
Step 3: Find the area
A= 15/2 – 4 = 15/2 – 8/2 = 7/2
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During rainy season, rate of evaporation decreases because (a) humidity is high in air. (b) humidity is low in air. (c) humidity does not change. (d) there is no effect of humidity.
(a) The humidity is high in air, therefore the rate of evaporation decreases. There is limit to moisture that the air can carry at a particular temperature. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
(a) The humidity is high in air, therefore the rate of evaporation decreases. There is limit to moisture that the air can carry at a particular temperature.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
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