As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "
Doppler's effect."
Now the general formula of the Doppler's effect is:
-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.
= Velocity of the receiver/observer relative to the medium = ?.
= Velocity of the source with respect to medium = 0 m/s.
= Frequency emitted from source = 400 Hz.
= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:
Solving above would give you,
= 6.8 m/s
The correct answer = 6.8m/s
<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
Answer:R=1607556m
θ=180degrees
Explanation:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Using analytical method :
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
Note that θ is in the second quadrant, so add 180
θ=180-7.9256*10^6=180degrees
Answer:
1.0125 x 10^19
Explanation:
current flowing through conductive wire= 9mA = 9 x 10^ -3 A
charge passing per 3 min
Q = It
= 9 x 10^ -3 x (3 x 60)
= 1.620 C
no of electrons in charge
Q = ne
1.620 = n x 1.6 x 10 ^ -19
n. = 1.0125 x 10 ^19
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that
Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then
Therefore, the coefficient of kinetic friction is approximately 0.0367
