Q out = Q in
Q mix = Q water
mcΔt sand + mcΔt water = mcΔt water
30.2 x 2.01 x (29.3-12.1) + 87.7 x 4.19 x (29.3-12.1)=m x 4.19 x (85.8-29.3)
1044.07+6320.36=m x 236.74
m = 31.11 g
When 2 electric charges are place close to each other they experience a force between them. This force may be repulsive or attractive depending on the type of charges involved.
The magnitude of this force depend on the quantity of individual charges and the distance between them.
The force is defined by the <em>Coulomb's law</em> states that: <em>The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.</em>
The formula for calculating the force between 2 electric charges is ;
F = (kq₁q₂)/d²
Where k is the proportionality constant known as Coulomb's constant,
q₁ and q₂ are the charges and
d is the distance between q₁ and q₂.
Explanation:
Given , F = 30N and mass m = 90kg
°•° F = ma
=> a = F/m
=> a = 30/90
=> a = 1/3m/s^2
Answer:
0.74 N/cm
Explanation:
The following data were obtained from the question:
Mass (m) = 3 Kg
Extention (e) = 40 cm
Spring constant (K) =?
Next, we shall determine the force exerted on the spring.
This can be obtained as follow:
Mass (m) = 3 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 3 × 9.8
F = 29.4 N
Finally, we shall determine the spring constant of the spring. This can be obtained as follow:
Extention (e) = 40 cm
Force (F) = 29.4 N
Spring constant (K) =?
F = Ke
29.4 = K × 40
Divide both side by 40
K = 29.4 / 40
K = 0.74 N/cm
Therefore, the spring constant of the spring is 0.74 N/cm
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).