If i thought about this properly the answer is b
Answer:
a. 4.9 m
Explanation:
To solve this problem we must take into account that power is defined as the relationship between the work and the time in which the work is done.
P = W/t
where:
P = power = 95 [W] (units of watts)
W = work [J] (units of Joules)
t = time = 6.2 [s]
We can clear the work from the previous equation.
W = P*t
W = 95*6.2 = 589 [J]
Now we know that the work is defined by the product of the force by the distance, therefore we can express the work done with the following equation.
W = F*d
where:
F = force = 120 [N] (units of Newtons)
d = distance [m]
d = W/F
d = 589/120
d = 4.9 [m]
I beileve the answer is B
Answer:
2.97795 Hz
Explanation:
v = Speed of sound in air = 343 m/s
= Relative speed between the speakers and the student = 1.12 m/s
= Actual frequency of sound = 456 Hz
Frequency of sound heard as the student moves away from one speaker
![f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz](https://tex.z-dn.net/?f=f_1%3Df%27%5Cdfrac%7Bv-v_r%7D%7Bv%7D%5C%5C%5CRightarrow%20f_1%3D456%5Cdfrac%7B343-1.12%7D%7B343%7D%5C%5C%5CRightarrow%20f_1%3D454.51102%5C%20Hz)
Frequency of sound heard as the student moves closer to the other speaker
![f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz](https://tex.z-dn.net/?f=f_2%3Df%27%5Cdfrac%7Bv%2Bv_r%7D%7Bv%7D%5C%5C%5CRightarrow%20f_2%3D456%5Cdfrac%7B343%2B1.12%7D%7B343%7D%5C%5C%5CRightarrow%20f_2%3D457.48897%5C%20Hz)
The difference in the frequencies is
![f_2-f_1=457.48897-454.51102=2.97795\ Hz](https://tex.z-dn.net/?f=f_2-f_1%3D457.48897-454.51102%3D2.97795%5C%20Hz)
The student hears 2.97795 Hz
I believe that it is true! :)