To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as
c = Propagation speed of waves in the medium
= Speed of the receiver relative to the medium
= Speed of the source relative to the medium
Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
Velocity of car
velocity of motor
Velocity of sound
Frequency emited by the source
Replacing we have that
Therefore the frequency that hear the motorcyclist is 601.7Hz
Answer:
θ=180°
Explanation:
The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.
Using the right hand rule for vector product, we can test the two possible cases:
- If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.
- If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.
Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.
