A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
1 answer:
The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."
The standard formula for the Doppler effect is:
-- (A)
Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.
Where,
g = Speed of sound = 340m/s.
= Velocity of the observer relative to the medium =?.
= Velocity of the source in relation to the medium = 0 m/s.
= Frequency emitted from the source = 400 Hz.
= Frequency recognized by the observer = 408 Hz.
Substituting the given values into equation (A) will yield:
Solving the above will result in,
= 6.8 m/s
The correct result = 6.8m/s
The standard formula for the Doppler effect is:
Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.
Where,
g = Speed of sound = 340m/s.
Substituting the given values into equation (A) will yield:
Solving the above will result in,
The correct result = 6.8m/s
You might be interested in
Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.
Then, we find the final velocity
The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.
Answer:
Explanation:
a) La fuerza neta que actúa sobre la caja en la dirección vertical es:
Fnet=Fg−f−Fp *sin45 °
aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.
Fnet=ma
ma=Fg−f−Fp *sin45 °
a=
=0.24 m/s²
Vf =Vi +at
=0.48+0.24*2
Vf=2.98 m/s
b)
Fnet=Fg−f−Fp *sin45 °
=Fg−0.516Fp−Fp *sin45 °
=30-1.273Fp
Fnet=0 (Ya que la velocidad es constante)
Fp=30/1.273
=23.56 N