A man stands one-third of the way between two walls and claps his hands, He hears two distinct echoes 1s apart. If the speed of sound is 330ms^-1 the distance between the wall is

Onesimus75

7 years ago

That means the distance is
d= 1/3
using
v= 2d/t
330 = 2d×1/3 / 1
330= 2d/3
330×3 = 2d
d= 990/2
d= 495

option E is correct

MikeFC

5 years ago

v=2d/t
but the man stands one third of his distance =⅓d
Therefore, v=2×⅓d/t
330=2×⅓d/1
330=2d/3
Making d subject formula
2d=330×2
d=990/2
d=495

Wonkanen

4 years ago

V=2x/t
330=(2*x*1/3)/1
330=(2x/3)/1
330=2x/3
Cross muliply
330*3=2x
990=2x
Divide both side by 2
X=495

3BYKE

4 years ago

the man was standing at one third of d(distance between two walls), that means that he is 2/3 d from the other wall, and the echo were distinct echoes (not same reflection)
so since the v is same, make v subject for the two echoes
first echo distance=1/3d, time =x,
for second echoe distance=2/3d, time =(x+1)seconds
from eqn v=2d/t
therefore (2(1/3d))/x =(2(2/3d))/x+1
after solving this u would get x =1
then sub 1 for x in the equation v=2d/t to get ur d, so 330=2(1/3d)/1, d =495

OsagieN

2 years ago

𝐱 + 𝐲 = 𝐝...........𝐲 = 𝟮𝐃/𝟯...........𝐱 = 𝐃/𝟯

𝐭₁ - 𝐭₂ = 𝟭

𝟮𝐲/𝐯 - 𝟮𝐱/𝐯 = 𝟭

𝟮/𝐯 (𝐲 - 𝐱) = 𝟭

𝟮/𝐯 (𝟮𝐃/𝟯 - 𝐃/𝟯) = 𝟭

𝟮/𝐯 (𝐃/𝟯) = 𝟭

𝟮𝐃 = 𝟯𝐯

𝐃 = 𝟯(𝟯𝟯𝟬)/𝟮 = 𝟵𝟵𝟬/𝟮 = 𝟰𝟵𝟱 𝗺𝗲𝘁𝗿𝗲𝘀........

Chukwuedomiracle

5 years ago

Yes it's possible, before reprinting