Answer:The wave patterns are transverse waves.
looks like a slope. The waves move up and down and spread out. The medium has particles that vibrate in a direction perpendicular to the direction of the propagation of wave.
Explanation:
Answer:i think they do i dont understand your question it does not make sence
Explanation:
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Incomplete question as time is missing.I have assumed some times here.The complete question is here
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Explanation:
Given data
Vi=10 m/s
S=70 m
(a) t₁=0.5 s
(b) t₂=1 s
(c) t₃=1.5 s
(d) t₄=2 s
(e) t₅=2.5 s
To find
Displacement S from t₁ to t₅
Velocity V from t₁ to t₅
Solution
According to kinematic equation of motion and given information conclude that v is given by

Also get the equation of displacement

These two formula are used to find velocity as well as displacement for time t₁ to t₅
For t₁=0.5 s

For t₂
For t₃

For t₄

For t₅

Answer:
15.64 KN
Explanation:
mass of the elevator cab with a single occupant= 2300 kg
acceleration relative to the cab
= 6.80 m/s^2
acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2
The acceleration of elevator cab relative to the ground 
now we can say that

=-6.80+ a_{eg}= -9.8
[tex]a_{eg}=-9.8+6.80=-3.8
The forces that act on elevator cab are tension and gravitational, applying newtons second law
T- mg= ma_{eg}
Then the tension in the cable is
T= 2300(-3.8)+2300×9.8= 15640 N= 15.64 KN
therefore tension in the string will be 15.64 KN