Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
<span>It's pretty easy problem once you set it up.
Earth------------P--------------Moon
"P" is where the gravitational forces from both bodies are acting equally on a mass m
Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon
You are correct to use that equation. If the gravitational forces are equal then
GMearth*m/Rep² = Gm*Mmoon/Rpm²
Mearth/Mmoon = Rep² / Rpm²
Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be
Rpm = Rem - Rep
Mearth / Mmoon = Rep² / (Rem - Rep)²
Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²
Everything is done now. The most complicated part now is the algebra,
so bear with me as we solve for Rep. I may skip some obvious or
too-long-to-type steps.
81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0
We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem
Obviously, point P cannot be 9/8 of the way to the moon because it'll be
beyond the moon. Therefore, the logical answer would be 9/10 the way
to the moon or B.
Edit: The great thing about this idealized 2-body problem, James, is
that it is disguised as a problem where you need to know a lot of values
but in reality, a lot of them cancel out once you do the math. Funny
thing is, I never saw this problem in physics during Freshman year. I
saw it orbital mechanics in my junior year in Aerospace Engineering. </span>
sylent_reality
· 8 years ago
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
Answer:
Horizontal component: 
Vertical component: 
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
For the vertical component, we have:
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.
Answer:
This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.
Explanation:
This is because the rubbing releases negative charges
