Answer:
I would say since it is still water the stream flow is 0km/h. Especially because he needs the same time upstream as He does down stream.
Answer:
1 : 2 cm/km or ![\frac{1}{2} \, \frac{cm}{km}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5C%2C%20%5Cfrac%7Bcm%7D%7Bkm%7D)
One cm represents 2 km.
Explanation:
If 3 cm represents 6 km on the earth, the ratio to consider is: 3 : 6 cm/km which is the same as:
1 : 2 cm/km or ![\frac{1}{2} \, \frac{cm}{km}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5C%2C%20%5Cfrac%7Bcm%7D%7Bkm%7D)
One cm represents 2 km.
Answer:
Explanation:
Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.
Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.
Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.
Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS
Answer:
Explanation:
Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.
Energy dissipates in 55Ω resistor is given by V²/R
Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.
So for 55ohms, using voltage divider rule
V=R1/(R1+R2) ×Vt
V=55/(55+140) ×70
V=19.74Volts is across the 55ohms resistor.
Then, energy loss will be
E=V²/R
E=19.74²/55
E=7.09J
7.09J of heat is dissipated by the 55ohms resistor
It will be cloudy and there will be rain.
Hope this helps