Answer:
4.9 m/s
Explanation:
Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the ball in this problem,
u = 0 (it starts from rest)
is the acceleration
s = 3 m is the distance covered
Solving for v,
What is the variable?
~<em>the</em><em> </em><em>price</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>variable</em><em>.</em>
What happens to demand?
~It'll go down. Since the price of snow blowers will increase then the quantity demanded will go down.
Hope this helps- have a good day bro cya)
B. The gravity acceleration is in the same direction as the force of gravity, and thus towards the centre of the earth
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- 
--- 
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1
The new force (F2) is
Substitute 4 for r2
Substitute
The new force is 1/4 of the previous force.
