Answer: 2. Solution A attains a higher temperature.
Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.
In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.
Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.
<em>We have a formula for such condition,</em>
.....................................(1)
where:
= temperature difference
- c= specific heat of the body
<u>Proving mathematically:</u>
<em>According to the given conditions</em>
- we have equal masses of two solutions A & B, i.e.

- equal heat is supplied to both the solutions, i.e.

- specific heat of solution A,

- specific heat of solution B,

&
are the change in temperatures of the respective solutions.
Now, putting the above values


Which proves that solution A attains a higher temperature than solution B.
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =
is located in the center of a spherical cavity of radius ,
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at
m due to shell
E1 = 

Electric field at
due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

Answer:
Explanation:
The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula
Intensity of sound = Power / 4π d²
= .535 / 4 x 3.14 x (27.3/2)²
= 2.286 x 10⁻⁴ J m⁻² s⁻¹
Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹
Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹
= 4.57 x 10⁻⁴J m⁻² s⁻¹
b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .
The total intensity of sound reaching him is as follows
0.535 / (4 π x18.15² ) + 0.535 / (4 π x9.15² )
= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴
= 6.38 x 10⁻⁴J m⁻² s⁻¹
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA