7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
Answer:
0.686 g of ice melts each second.
Solution:
As per the question:
Cross-sectional Area of the Copper Rod, A = ![11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}](https://tex.z-dn.net/?f=11.6%5C%20cm%5E%7B2%7D%20%3D%2011.6%5Ctimes%2010%5E%7B-%204%7D%5C%20m%5E%7B2%7D)
Length of the rod, L = 19.6 cm = 0.196 m
Thermal conductivity of Copper, K = ![390\ W/m.^{\circ}C](https://tex.z-dn.net/?f=390%5C%20W%2Fm.%5E%7B%5Ccirc%7DC)
Conduction of heat from the rod per second is given by:
![q = \frac{KA\Delta T}{L}](https://tex.z-dn.net/?f=q%20%3D%20%5Cfrac%7BKA%5CDelta%20T%7D%7BL%7D)
where
= temperature difference between the two ends of the rod.
Thus
![q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s](https://tex.z-dn.net/?f=q%20%3D%20%5Cfrac%7B390%5Ctimes%2011.6%5Ctimes%2010%5E%7B-%204%7D%5Ctimes%20100%7D%7B0.196%7D%20%3D%20228.48%5C%20J%2Fs)
Now,
To calculate the mass, M of the ice melted per sec:
![M = \frac{q}{L_{w}}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bq%7D%7BL_%7Bw%7D%7D)
where
= Latent heat of fusion of water = 333 kJ/kg
![M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B228.48%7D%7B333%5Ctimes%2010%5E%7B3%7D%7D%20%3D%206.86%5Ctimes%2010%5E%7B-%204%7D%5C%20kg%20%3D%200.686%5C%20g)
Speed is equal to distance traveled divided by the time. So it's 3.5 m/s
Answer:
Did you ever get the answer?
Explanation:
Answer:6 joules
Explanation:
Mass(m)=3kg
Velocity(v)=2m/s
Kinetic energy=0.5 x m x v^2
Kinetic energy=0.5 x 3 x 2^2
Kinetic energy=0.5 x 3 x 2 x 2
Kinetic energy=6