Answer:
There are actually three, Kepler’s laws that is, of planetary motion: 1) every planet’s orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit. As it’s the third which is most often used, Kepler’s law usually means Kepler’s third law (of planetary motion).
Explanation:
Kepler's third law would tell us that Halley's comet has an average distance much greater than that of the Earth. However, there is a time in Halley's comet's orbit that brings it closer to the Sun than the Earth. Kepler's third law is a mathematical relation between a planet's period and its average distance.
Answer:
t = 2s
Explanation:
When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)
Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.
With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.
So at t = 2s, we have an instantaneous acceleration of -2m/s².
Answer:
2.44 m/s due East
Explanation:
From the question given above, the following data were obtained:
Mass of 1st car (m₁) = 4 Kg
Velocity of 1st car (u₁) = 3 m/s
Mass of 2nd car (m₂) = 5 Kg
Velocity of 2nd car (u₂) = 2 m/s
Final velocity (v) =?
The final velocity can be obtained as follow:
v(m₁ + m₂) = m₁u₁ + m₂u₂
v(4 + 5) = (4×3) + (5×2)
9v = 12 + 10
9v = 22
Divide both side by 9
v = 22/9
v = 2.44 m/s
Thus, the final velocity is 2.44 m/s.
Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.
