(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.
The lower the frequency the lower the pitch sound.
Explanation:
Initial energy = final energy + work done by friction
PE = PE + KE + W
mgH = mgh + 1/2 mv² + W
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000
v = 22.1 m/s
Without friction:
PE = PE + KE
mgH = mgh + 1/2 mv²
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²
v = 23.4 m/s
Answer:
Control of air–fuel ratio
Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).
Explanation:
