Q1: SPM2004/P1
The diagram shows an ultrasonic waves
transmitted from a boat to the seabed to
determine the depth, D, of the sea. The
speed of the ultrasonic waves in water is
1 500 m s ‘.The echo of the waves is received
2.0 s after the transmission.
transmitted from a boat to the seabed to
determine the depth, D, of the sea. The
speed of the ultrasonic waves in water is
1 500 m s ‘.The echo of the waves is received
2.0 s after the transmission.
What is the value of D?
A 375m
B 750m
C 1500m
D 3000m
E 6000m
ans:C
Discussion:
thus D= (1500 x 2)/ 2
Q2: SPM2003/P1
Ahmad shouts in front of a high wall. He hears the echo of his voice 1.2 seconds
later.The velocity of sound in air is 340 m/s.What is the distance between Ahmad and
the wall?
A 204m
B 283m
C 340m
U 408m
E 816m
ans:A
Discussion:
.. like in 2004 almost similar pattern of question came out in 2003. U can apply the same formula 2D = v x t ( D=distance, v=velocity, t=total time).
Q3: SPM2006/P1
Again..the same pattern came out in 2006
Diagram 21 shows a submarine transmitting ultrasonic waves directed at a big rock on the sea bed. After 10 seconds, the submarine detects the reflected wave.
Calculate the distance of the submarine from
the big rock.
[Velocity of ultrasonic wave =
1 560m/s]
A. 3.9 km
B. 7.8 km
C. 15.6 km
D. 31.2 km
E. 156.6 km
answer:
try this out first.. u can leave a comment to know the answer :).
2 comments:
for Question 3, the answer is C=15.6km, right??
answer is B.
u must consider time taken 10s is the duration after it bounced back to the submarine, that means the distance u've got 15.6km is traveled from submarine to the rock & back to submarine. Assume distance frm submarine to rock is D,after bounce back total distance traveled by the wave is 2D.
velocity = total distance /time
v = 2D / t
2D = vt=1560 x 10=15600m/s
D = (vt)/2
= 15600/2 =7800m/s
=7.8km/s
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