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:
271.862 N/m
Explanation:
From Hook's Law,
mgh = 1/2ke²............... Equation 1
Where
m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.
Making k the subject of the equation,
k =2mgh/k²....................... Equation 2
Note: The potential energy of the ball is equal to the elastic potential energy of the spring.
Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m
Substitute into equation 2
k = 2(0.0603)(9.8)(0.537)/0.048317²
k = 0.6346696/0.0023345
k = 271.862 N/m
Answer:
C. Equals
Explanation:
Law of reflection Equals the angle of incidence
Answer:
The level of the root beer is dropping at a rate of 0.08603 cm/s.
Explanation:
The volume of the cone is :
Where, V is the volume of the cone
r is the radius of the cone
h is the height of the cone
The ratio of the radius and the height remains constant in overall the cone.
Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm
h = 13 cm
r / h = 5 / 13
r = {5 / 13} h
Also differentiating the expression of volume w.r.t. time as:
Given: 
= -4 cm³/sec (negative sign to show leaving)
h = 10 cm
So,
<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>
