1.a source of ignition
2.fuel
3.oxygen
If wave exhibits reinforcement, the component waves must be in phase with each other. Components waves combine to form a resultant with the same wavelength but the amplitude which is greater than the amplitude of either of the individual component waves, and this happens in constructive interference.
In phase the features of the two waves they completely match at zero degrees.
Answer:
Description 1 matches C.
Description 2 matches A.
Description 3 matches D.
Description 4 matches B.
Explanation:
Description 1 matches C. From graph C, we can see that the <em>distance remains constant at a particular value</em>. Hence, the car is <em>not traveling</em> and hence it is <em>stopped.</em>
Description 2 matches A. <em>Speed is determined by the gradient of a distance-time graph</em>. A depicts a <em>linear graph</em> which tells us that the <em>gradient is constant </em>and hence the car is at <em>constant speed</em>. The fact that the<em> distance is increasing</em> shows that the<em> car is moving forward</em>.
Description 3 matches D. Graph of D becomes <em>less steeper as time progresses</em>. This tells us that the <em>gradient of the graph is decreasing </em>and hence the <em>speed of the car is decreasing.</em>
<em />
Description 4 matches B. Graph B shows <em>distance being decreased.</em> This tells us that the <em>car is coming back</em>.
35m/s is meters per second.
35x35=1225
it will have gone a total of 1225 meters on thirty five seconds
Answer:
<u>NORMAL DRIVER: </u>d = 73.3 ft
<u>DRUNK DRIVER: </u>d = 172.3
Explanation:
<u>NORMAL DRIVER:</u>
Distance covered in initial 0.75s = 0.75s *44 = 33ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 33 + 40.3 = 73.3 ft
<u>DRUNK DRIVER:</u>
Distance covered in initial 3s = 3s *44 = 132 ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 132 + 40.3 = 172.3 ft