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:
14.7 m/s.
Explanation:
From the question given above, the following data were obtained:
Time (t) = 1.5 s
Acceleration due to gravity (g) = 9.8 m/s².
Height = 11.025 m
Final velocity (v) = 0 m/s
Initial velocity (u) =?
We, can obtain the initial velocity of the penny as follow:
H = ½(v + u) t
11.025 = ½ (0 + u) × 1.5
11.025 = ½ × u × 1.5
11.025 = u × 0.75
Divide both side by 0.75
u = 11.025/0.75
u = 14.7 m/s
Therefore, the penny was travelling at 14.7 m/s before hitting the ground.
The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
Answer:
Do find the answer in the attachment herein.
Explanation:
From the attached diagram:
I. Activation energy = Activated complex - ∆H(reactants)
Activation energy = 162-140 = 22Kj.
II. ∆H(reaction) = ∆H(products) - ∆H(reactants)
∆H(reaction) = 37 - 140 = -103Kj.
Answer:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j