Heat capacity of A is three times that of B
and the initial temperature for A is twice that of B
with TA = 2 TB
Let T denote the final temperature of the system.
The heat lost by A equals the heat gained by B:
mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)
which simplifies to heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)
resulting in 3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB).
This leads us to the equation: 3 TA - 3 T = T - TB
which rearranges to yield 6 TB + TB = 4 T
thus giving us T = 1.75 TB
Answer:
The resulting value is 
.
Explanation:
The question specifies that
The wavelength is 
.
The velocity is 
.
The mass of the electron is 
.
The energy of the incoming light is typically depicted mathematically as
.
Here, c represents the speed of light with the value 
.
h stands for Planck's constant with a value of 
.
Thus,
=> 
.
Typically, kinetic energy is represented as
=> 
.
=> 
.
The ionization energy is generally expressed mathematically as
=> .
THE GREEN HOSE: Define the (x,y) coordinates at a height of 4 feet, which corresponds to where Majra holds the green hose. This indicates the equation for the green hose takes the form y = a(x - h)² + 4. Water from the hose lands on the ground 10 feet away from Majra, thus y(10) = -4. Given that the curve passes through (0,0), this leads to ah² + 4 = 0; therefore, ah² = -4. To satisfy the previous equation, we find a(10 - h)² + 4 = -4, simplifying to a(10 - h)² = -8. Dividing (3) by (4) gives a ratio of h²/(10-h)² = 1/2, leading to 2h² = (10 - h)² = 100 - 20h + h², and resolving yields h² + 20h - 100 = 0. Applying the quadratic formula, we get x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142. We discard the negative solution. The vertex locates at (4.142, 4). From (3), we deduce a = -4/4.142² = -0.2332, leading to the equation for the green hose: y = 0.2332(x - 4.142)² + 4. THE RED HOSE: The vertex of the red hose is positioned at (3,7), represented by the equation y = -(x-3)² + 7. A graph depicting y(x) for both hoses is included in the attached figure. Answers: a. The red hose throws water higher. b. The green hose's equation is y = -0.2332(x - 4.124)² + 4, starting at a height of 4 feet. c. The feasible domain for the green hose is between 0 ≤ x ≤ 10 feet, with the corresponding range being -4 ≤ y ≤ 4 feet.
Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
In this scenario, the principles of momentum conservation can be applied since there are no external forces acting on the system. Consequently, the conservation of momentum principle is applicable here. After the bird lands on it, both the bird and the bark will have a unified final speed. Thus, this final speed will be 1 m/s.
