The distance covered is 25.9 m.
<h3>How deep is the cave?</h3>
We know that the speed of sound refers to the speed with which an sound moves in an object.
Given that;
speed of sound = 345m/s
Time taken = 0.15s
We know that;
v = 2d/t
v = speed of sound
d = distance
t = time taken
vt = 2d
d = vt/2
d = 345m/s * 0.15s/2
d = 25.9 m
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Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
Explanation:
Below is an attachment containing the solution.
The speed of A and B immediately after collision is 5.28m/s
<u>Explanation:</u>
Mass of A is 6275kg
Speed of A is 6.5m/s
Mass of B is 5155kg
Speed of B is 3.8m/s
Track is frictionless.
A and B stick together.
speed of attached A and B = ?
mₐsₐ + mᵇsᵇ = (mₐ + mb) s

Therefore, The speed of A and B immediately after collision is 5.28m/s