Answer:
The centripetal acceleration will be "21.785 m/s²".
Explanation:
The given values are:
Time,
t = 0.85 seconds
Length of rope,
r = 0.40 m
Mass of ball,
m = 0.80 kg
As we know,
⇒
On substituting the values, we get
⇒
⇒
⇒
The centripetal acceleration will be:
⇒
⇒
⇒
⇒
Answer:
Explanation:
A projectile's vertical velocity at the top of its flight path must be zero, therefore the speed at the top of the trajectory must be referring to the object's horizontal velocity. Since there are exactly zero horizontal forces acting on the object, its horizontal velocity will remain the exact same throughout the entire launch. Thus, the horizontal component of the initial launch must be 45 meters per second, and we have the following equation using basic trigonometry for a right triangle:
Let be the unknown angle above the horizontal.
Gravitational Potential of an object with mass m, and height of h metre is :
Now, if the mass of object is 2100 kg, height of 18 m is placed in there, then it's gravitational potential energy is ~
This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
=======================================
If you make a list of the distances and directions, and ignore the times,
you find these:
4 - west, (3 + 1) - east . . . . . zero in the east/west direction
1.5 - north, 1.5 - south . . . . . zero in the north/south direction
This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.
Average velocity = (distance between start point and end point) / (time)
IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time. But it asks for VELOCITY, and <u>that</u> only involves the straight distance between the start point and the end point, regardless of the route taken in between.
The jogger ended up exactly where he started. The distance between start and end points was zero. Average velocity is (zero) / (time) . And that fraction is going to be <em><u>Zero</u></em>, no matter how long or how short the trip was, and no matter how much time it took.