To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,
The electric field is
Here,
k = Coulomb's Constant
Q = Charge
R = Distance (At this case from the center of mass of the earth to the surface)
Rearranging to find the charge,
Replacing,
Since the electric field is directed towards the center of earth, the charge is negative.
PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore
Replacing,
Solving for q,
PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is
PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.
Answer:true
Explanation:
Displacement is the vector representation of a change in position. It is path independent and is equivalent to the straight line distance between the start and end locations. Distance is a scalar quantity that reflects the path traveled.
Thes are all correct :<span>2.Scientists observe natural events occurring around them. They question these observations and investigate them.
3.All scientists follow one scientific method of investigation
4.If a hypothesis is not verified by the results from the scientific method, scientists may either redo the process or create a new hypothesis.</span>
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)
