Answer:
C.) The slinky particles move up and down
Explanation:
<u>Transverse Wave</u>-
<em>A wave that has a disturbance perpendicular to the wave motion</em>
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<em>Hello! This is the correct answer! Have a blessed day! :)</em>
<em>If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!</em>
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The volume flow rates for ∆P is 6.81m³/s .
<h3>What is pressure?</h3>
The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.
According to bernaulli's theorem theorem
P+1/2pV²+pgy = constant
where p fluid density
g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.
As there are two tubes then the height of tube 1 is equal to height of tube two .
P1-P2=1/2p(Vd²-Vl²)
The flow rate of liquid is A1V1=A2V2 .
rest is attached in image
to learn more about Pressure click here brainly.com/question/12971272
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Answer:
2.71 m
Explanation:
Force is the product of mass and acceleration
F=m*a
Work done is the product of force and distance
Work done=F*d
In this case;
F= 35 N
Work done = 95 J
95 =35 * d
95 /35 = d
2.71 m= d
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671
Answer:
When you look at a simple koi pond you can find Koi (the secondary consumer) that feeds off of the zooplankton (first consumer), they eat the phytoplankton (producers). All in a simple food chain
Explanation:
Basically, Koi eat the little animal plankton (zooplankton) that then eats the plant plankton (phytoplankton) that can only end when a part of that habitat is removed. If you got rid of the plant plankton then the whole chain would collapse and most likely die.