Answer:
it appears to be farther away than it actually is, and therefore smaller then the object itself.
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer:it’s is part of the cell theory because they where studying cells and to see it you need a microscope
Explanation:basically in the answer area
•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
