How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
1 answer:
You might be interested in
Answer:
0.872<em>m/s</em>
Explanation:
Tangential velocity is given by the formula,
In the question given,
radius= 25meters
time= 180secs
pie= 3.14
number of laps= 1
The magnitude of tangential velocity equals;
<em>v </em>= 157<em>m</em>/180<em>secs</em>
Therefore, the magnitude of the tangential velocity
=0.872<em>m/secs</em>
Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
<u>P = 2439.5 W = 2.439 KW</u>