Answer:
As waves get closer to a beach they decrease in height.
Explanation:
As a wave crest approaches the shoreline, it is usual that one end of the line is closer to the shoreline than the other. The implication of this is that the energy in a wave is also spread over quite a larger area,this in turn reduces the height of the waves. In other words, refraction often makes waves smaller.
Waves are caused by wind. Wave height in the open ocean is determined by three factors. The greater the wind speed the larger the waves. The greater the duration of the wind (or storm) the larger the waves. The greater the fetch (area over which the wind is blowing - size of storm) the larger the waves.
The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.
<h3>
What is the acceleration?</h3>
Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.
Thus;
v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time taken
v - u/t = a
a = 6 - 0/3
= 2 m/s^2
Again;
v - u/t = a
a = 0 - 6/5
a = - 1.2m/s^2
Learn more about acceleration:brainly.com/question/12550364
#SPJ1
<h3 />
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
It will take 2.68 minutes for them to reach each other.
Explanation:
We use the two following kinematic equations, making the final position the same (for the moment they meet each other):
locomotive 1 --> 
locomotive 2 --> 
we make the two xf equal, and solve for the time (t) using v = 95 km/h:
converting the hours into minutes by multiplying this value times 60;
t = 2.68 minutes
<span>A transverse wave is characterized by peaks and dips.</span>
