Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.
Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.
<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
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<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,
We have
Then the total work is
Answer:
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Explanation:
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Answer:
Motion maps are used to illustrate the direction and position of an object. Using the motion map, the description of the object position and velocity is as follows:
The object starts its movement from the origin with a large velocity, before moving back to the origin with a smaller velocity. It stops for 1 second in the origin, then moves away with a larger velocity, Finally, it moves back towards the origin with a smaller velocity.
See attachment for the motion map, where the number on each arrow in the map, represents the position of the object.
Note that; the long arrow means large velocity while the short arrow means small velocity
Next, we analyze the direction and position using the arrows
The first arrow shows that the object starts from the origin with a large velocity
The direction and length of the second arrow show that, the object then returned to the origin with a smaller velocity.
There is a dot in front of the second arrow. This dot indicates that the object stops for one second.
The third arrow means that, the object moved from the origin with a larger velocity
The direction and position of the fourth and fifth arrows indicate that the object then moves towards the origin, with a smaller velocity.
Explanation:
