Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
Given Information:
Frequency of horn = f₀ = 440 Hz
Speed of sound = v = 330 m/s
Speed of bus = v₀ = 20 m/s
Answer:
Case 1. When the bus is crossing the student = 440 Hz
Case 2. When the bus is approaching the student = 414.9 Hz
Case 3. When the bus is moving away from the student = 468.4 Hz
Explanation:
There are 3 cases in this scenario:
Case 1. When the bus is crossing the student
Case 2. When the bus is approaching the student
Case 3. When the bus is moving away from the student
Let us explore each case:
Case 1. When the bus is crossing the student:
Student will hear the same frequency emitted by the horn that is 440 Hz.
f = 440 Hz
Case 2. When the bus is approaching the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330+20 )
f = 440 ( 330/ 350 )
f = 440 ( 0.943 )
f = 414.9 Hz
Case 3. When the bus is moving away from the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330-20 )
f = 440 ( 330/ 310 )
f = 440 ( 1.0645 )
f = 468.4 Hz
Answer:
3rd class lever
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Answer:
m = 9795.9 kg
Explanation:
v = 35 m/s
KE = 6,000,000 J
Plug those values into the following equation:
6,000,000 J = (1/2)(35^2)m
---> m = 9795.9 kg
