Explanation:
33. The 1.5kg owl is now soaring at 20m/s. What is the owl’s KE?
a. Step 1: Formula <u>½mv²</u>
b. Step 2: Data m = <u>1</u><u>.</u><u>5</u><u> </u><u>kg</u>, v = <u>2</u><u>0</u><u> </u><u>m</u><u>/</u><u>s</u>
c. Step 3: Solve
KE = (1/2)(<u>1</u><u>.</u><u>5</u>)(<u>2</u><u>0</u>)² = <u>3</u><u>0</u><u>0</u><u> </u><u>J</u>
Anyways the answer is x=2
Answer: Got It!
<em>Explanation: </em>let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
0 0
Answer:
a) q_1=q_2= 7.42*10^-7 C
b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C
Explanation:
Given:
F_e = 0.220 N
separation between spheres r = 0.15 m
Electrostatic constant k = 8.99*10^9
Find: charge on each sphere
part a
q_1 = q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_1^2 = F_e*r^2/k
q_1=q_2= sqrt (F_e*r^2/k)
Plug in the values and evaluate:
q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)
q_1=q_2= 7.42*10^-7 C
part b
q_1 = 4*q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_2^2 = F_e*r^2/4*k
q_2= sqrt (F_e*r^2/4*k)
Plug in the values and evaluate:
q_2= sqrt (0.22*0.15^2/4*8.99*10^9)
q_2= 3.7102*10^-7 C
q_1 = 14.8*10^-7 C