Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.
1 answer:
You might be interested in
Answer:
a) a = g / 3
b) x (3.0) = 14.7 m
c) m (3.0) = 29.4 g
Explanation:
Given:-
- The following differential equation for (x) the distance a rain drop has fallen has the form:
- Where, v = Speed of the raindrop
- Proposed solution to given ODE:
v = a*t
Where, a = acceleration of raindrop
Find:-
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
Solution:-
- We know that acceleration (a) is the first derivative of velocity (v):
a = dv / dt ... Eq 1
- Similarly, we know that velocity (v) is the first derivative of displacement (x):
v = dx / dt , v = a*t ... proposed solution (Eq 2)
v .dt = dx = a*t . dt
- integrate both sides:
∫a*t . dt = ∫dt
x = 0.5*a*t^2 ... Eq 3
- Substitute Eq1 , 2 , 3 into the given ODE:
0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2
= 1.5 a^2 t^2
a = g / 3
- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:
x (t) = 0.5*a*t^2
x (t = 3.0) = 0.5*9.81*3^2 / 3
x (3.0) = 14.7 m
- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:
m (t) = k*x (t)
m (3.0) = 2.00*x(3.0)
m (3.0) = 2.00*14.7
m (3.0) = 29.4 g
A. Formula: F=ma or F/m=a
10,000N/1,267kg≈7.9m/
B. Formula: a= and s=d/t
speed= 394.6/15
s=26.3m/s
a=
a=1.75m/
C. 7.9-1.75=difference of 6.15m/
D. The force that most likely caused this difference is friction forces