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 :).
for Question 3, the answer is C=15.6km, right??
ReplyDeleteanswer is B.
ReplyDeleteu 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