A wave produced by a door bell?
There is a sound wave yes.
Answer:
a) the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Explanation:
Given that;
Gravitational acceleration g = 9.81 m/s²
Mass m = 5 kg
Extension of the spring X = 50 mm = 0.05 m
Spring constant k = ?
we know that;
mg = kX
5 × 9.81 = k(0.05)
k = 981 N/m
a)
Given that; Acceleration of the elevator a = 2 m/s² upwards
Extension of the spring in this situation = X1
Force exerted by the spring = F
we know that;
ma = F - mg
ma = kX1 - mg
we substitute
5 × 2 = 981 × X1 - (5 ×9.81 )
X1 = 0.06019 m
X1 = 60.19 mm
Therefore the spring will stretch 60.19 mm
with the same box attached as it accelerates upwards
B)
Acceleration of the elevator = a
The spring is relaxed i.e, it is not exerting any force on the box.
Only the weight force of the box is exerted on the box.
ma = mg
a = g
a = 9.81 m/s² downwards.
Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²
Really long we’ll not long but far in distance
Answer: because ν = velocity/λ where ν and λ are the frequency and wavelegth of the wave.
Explanation: In order to explain this problem we have to consider the relationship between frequency and wavelengths which are related by the velocity of the wave as follows ν*λ=v where ν and λ are the frequency and wavelegth of the wave. These parameters have an inverse proportionality.
Then, ν = velocity/λ
