<span>gravitational force = 0.1 mg-wt. = 0.1 * 10^-6 * 9.8 N = Gm1m2/r^2
m1 = 40 kg m2 =15 kg and r = 0.2 m
Put in and find G</span>
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).
We can answer this using one of the equations of linear
motion:
v = d / t
where:
v = velocity
d = distance
t = time
<span>In the problem, we are asked to find for the time in
which Driver B will catch up to Driver A. Therefore, find the time when dA = dB. Rearranging the
equation and equation dA and dB will result in:</span>
<span>vA * tA = vB * tB
---> 1</span>
It was given that:
vA = 68 mph
tA = tB + 3 (since person A was travelling 3 hours
earlier)
vB = 85 mph
tB = unknown
Substituting into equation 1:
68 * (tB + 3) = 85 * tB
68 tB + 204 = 85 tB
tB = 12 hrs
Therefore driver B would catch up to driver A after 12
hrs.
<span> </span>
Answer:
B). to the right
Explanation:
Since the direction of magnetic field is into the page
So here we know that
now the velocity is from bottom to top
so we have
now the force on the moving charge is given as
now we have
so force will be towards Right
