Answer: The force does not change.
Explanation:
The force between two charges q₁ and q₂ is:
F = k*(q₁*q₂)/r^2
where:
k is a constant.
r is the distance between the charges.
Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.
If we also increase the distance between the charges two times, the new distance will be 2*r
Then the new force between them is:
F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2
This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.
Use equations of motion to find the velocity just before it hits the floor:
<span>Vf^2 = Vi^2 + 2gx </span>
<span>Final velocity = 4.42m/s </span>
<span>Impulse is change in momentum so: </span>
<span>m(Vf - Vi) = 0.05(0 - 4.42) </span>
<span>= - 0.221 kg.m/s
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Answer:
The object will not move.
Explanation:
If nothing pushes against it it will not move. If its not on a slant it will not move.
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
20 ms¯¹
Explanation:
3. Determination of the final velocity
From the question given above, the following data were obtained:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
Acceleration is simply defined as the change in velocity per unit time.
Mathematically, it can be expressed as:
Acceleration (a) = final velocity – Initial velocity / time
a = v – u / t
With the above formula, we can obtain the final velocity of the car as follow:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
a = v – u / t
5 = v – 0 / 4
5 = v / 4
Cross multiply
v = 5 × 4
v = 20 ms¯¹
Thus, the final velocity of the car is 20 ms¯¹
