The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
Answer:
Energy stored, E = 0.072 J
Explanation:
Given that,
Capacitance, 
Capacitance, 
These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts
We know that in parallel combination of capacitor, the equivalent capacitance is given by :

The energy stored in the capacitor is given by :

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.
Answer: w = 200N
Explanation:
w = mg
m = 20kg
g = 10m/s^ (approximately)
w = 20kg . 10m/s^
w = 200N
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC.
Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules
Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used
q=mCΔT
11466.58 Joules = 65.1g x 4.18 J / g C x ΔT
11466.58/(65.1gx4.18)=ΔT
ΔT=42.14oC
So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>