Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem

Put the value into the formula


We need to calculate the magnitude of the charge q₃
Using formula of net force

Put the value into the formula






Hence, The value of charge q₃ is 40.46 μC.
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
It must exist in p<span>lasma state.</span>
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²