<span>82.0 kg
I am going to assume that there is a typo for the number of joules of energy. Doing a google search for this exact question showed this question multiple times with a value of 4942 joules which makes sense given how close the "o" key is to the "9" key. Because of this, I will assume that the correct value for the number of joules is 4942. With that in mind, here's the solution.
The gravitational potential energy is expressed as the mass multiplied by the height, multiplied by the local gravitational acceleration. So:
E = MHA
Solving for M, the substituting the known values and calculating gives:
E = MHA
E/(HA) = M
4942/(6.15*9.8) = M
4942/60.27 = M
81.99767712 = M
Rounding to 3 significant figures gives 82.0 kg</span>
Exothermic is the right kind of reaction.
Answer:
C. A rocket traveling to the Moon
Answer:
Explanation:
The magnitude of the acceleration makes an angle of 30° with the tangential velocity.
Resolving the acceleration to tangential and radial acceleration
at = aCos30 = √3a/2
ar = aSin30 = ½a
a = 2•ar
Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:
at = Rα
Then, α = at/R
since at = √3a/2
Then, α = √3 at/2R, equation 1
The radial acceleration is given as
ar = ω²R
Note that, at² + ar² = a²
at = √(a²-ar²)
Back to equation 1
α = √3 at/2R
α = √3√(a²-ar²)/2R
α = √3√(a²-(w²R)²)/2R
α = √3(a²-w⁴R²) / 2R
Also, a = 2•ar = 2w²R
Then,
α = √3((2w²R)²-w⁴R²) / 2R
α = √3(4w⁴R²-w⁴R²) / 2R
α = √3(3w⁴R²) / 2R
α = √9w⁴R² / 2R
α = 3w²R / 2R
α = 3w²/2
