Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.
IMA = Ideal Mechanical Advantage
First class lever = > F1 * x2 = F2 * x1
Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2
=> F1/F2 = x1 /x2
IMA = F1/F2 = x1/x2
Now you can see the effects of changing F1, F2, x1 and x2.
If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.
If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.
If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.
If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.
Answer: Increase the length between the applied effort and the pivot.
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= 
|v|= 
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
Answer:
v = 7.67 m/s
Explanation:
Given data:
horizontal distance 11.98 m
Acceleration due to gravity 9.8 m/s^2
Assuming initial velocity is zero
we know that
solving for t
we have
substituing all value for time t
t = 1.56 s
we know that speed is given as
v = 7.67 m/s
Answer:
Open- closed
Explanation:
It has Open-closed configuration of its end
