Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Answer: a) the force will be repulsive
b) the ratio of the new force to the old force will be 2
c) O
Explanation:
a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.
b) initial force will be -q(-Q)/d2
Adding extra charge -Q will cause change on B to become -2Q
The new force will be - 2Q(-q)/d2
Dividing new force by old force will give 2
C) if B is neutralized, the net charge becomes 0 and there will be no force on it.
Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the change the object's motion.
I’m gonna need more information about this
