Answer:

Explanation:
Given:
- mass of water,

- initial temperature of water,

- initial temperature of pan,

- mass of pan,

- mass of water evapourated,

- specific heat of water,

- specific heat of aluminium pan,

- latent heat of vapourization,

<u>Using the equation of heat:</u>
<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>



Answer:
The heat capacity for the second process is 15 J/K.
Explanation:
Given that,
Work = 100 J
Change temperature = 5 k
For adiabatic process,
The heat energy always same.


We need to calculate the number of moles and specific heat
Using formula of heat


Put the value into the formula


We need to calculate the heat
Using formula of heat

Put the value into the formula


We need to calculate the heat capacity for the second process
Using formula of heat

Put the value into the formula



Hence, The heat capacity for the second process is 15 J/K.
Answer:
F = 0.1575 N
Explanation:
When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.
In this moment then
Sphere one has a charge = Q/2
Sphere three has a charge = Q/2
Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.
How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q
Sphere two has a charge = 3/4Q
Sphere three has a charge = 3/4Q
The electrostatic force that acts on sphere 2 due to sphere 1 is:
F = 
F= 
how
= 0.42
Then
F = 
F = 0.1575 N
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg