KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s
Acceleration = (change in speed) / (time for the change)
Change in speed = (speed at the end) minus (speed at the beginning.
The cart's acceleration is
(0 - 2 m/s) / (0.3 sec)
= ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .
Newton's second law of motion says
Force = (mass) x (acceleration) .
For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)
= ( - 1.5 x 20/3 ) (kg-m/s²)
<span> = </span>- 10 newtons .
<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that
Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by
PART A) Replacing the values to find a, we have
Therefore the semimajor axis is 
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by
Answer:
8.5 Ω
Explanation:
La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
La fórmula de la resistencia (R) viene dada por:
R = ρL/A
Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.
Dado que:
L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm
Sustituyendo da:
R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶
R = 8.5 Ω
You've already told us the speed in ft/s . It's right there in the question. You said that light travels about 982,080,000 ft/s.
We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.
In scientific notation, it's written . . . <em>9.8208 x 10⁸ ft/s .</em>
We don't know where you were going with the number of seconds in a year. But to answer the question that you eventually asked, it turned out that we don't even need it.
