1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)
sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N
use formula: sin or cos
i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.
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:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
But the force F
But the electric field intensity due to a point charge Q at a distance r meters away is given by
<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>
Answer:
Explanation:
given data
density of current sheet = 0.40 A/m
length a = 0.27 m
width b = 0.63 m
For infinite sheet, magnetic field is given as
magnetic flux is given as
To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer 
is more than the frequency emitted by the source. The expression to find the frequency received by the person is,
= Frequency of the source
= Speed of sound
= Speed of source
The velocity of the ambulance is
Replacing at the expression to frequency of observer we have,
Therefore the frequency receive by observer is 878Hz
