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:
Slope = 0.50.
Explanation:
Below is an attachment containing the solution to the question.
We can use the ideal gas
equation which is expressed as PV = nRT. At a constant volume and number of
moles of the gas the ratio of T and P is equal to some constant. At
another set of condition, the constant is still the same. Calculations are as
follows:
T1/P1 = T2/P2
T2 = P2 x T1 / P1
T2 = 225 x 300 / 198
<span>T2 = 340.91 K</span>
Answer:
3. 1.8A
Explanation:
Given the following data;
Quantity of charge, Q = 650C
Time = 6 minutes to seconds = 6 * 60 = 360 seconds.
To find the current l;
Quantity of charge = current * time
Substituting into the equation, we have;
650 = current * 360
Current = 650/360
Current = 1.8 Amperes
