Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
Answer:
Yes. Inertia keeps the speed maintained though my feet leave the ground.
Explanation:
Inertia is the resistance to the change in position of any object this means this resistance will keep me traveling at 30 km/s relative to the sun. If the person wants to change the position we apply force to do that because inertia is opposing us to not do that. We are always traveling with 30km/s relative to sun due to inertia.
الملك - الإتاوات - المحاكم - المجلس الملكي - القانون الروماني
The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass,
= 72 kg
speed,
= 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X
+
X
= (
+
) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X 
v = 417.6 + 
v = 5.87 m/s
The final velocity is 5.87 m/s