The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.
That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)
If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)
At that point, since it has no more upward speed, it can't go any higher. Right ?
(crickets . . .)
Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
Answer:
Instantaneous speed means speed at any instant
that means Speed is changing with time
You know speed is distance/time
So that means distance is also changing with time
So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt
So, we take speed = ds/dt
ds = infinitesimal small distance
dt = infinitesimal small time
As its ratio is equal to speed at any instant
Note : We are taking infinitesimal small distance
But :) we are taking infinitesimal small time also
As you know if denominator is small fraction is large So fraction always give large value
So it's not O ( this makes confuse to most of students)
So, thanks
Good question
Keep thinking like this :)
<span>Average velocity can be calculated by determining the total displacement divided by the total time of travel. The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point. Average velocity is different from average speed because it considers the direction of travel and the overall change in position.</span>
