Answer:
b. The velocity remains constant
Explanation:
Answer:
It increases by four times the initial amplitude.
Explanation:
The amplitude of a sounde is inversely proportional to the square of the distance between the listener and the sound source:
![A \sim \frac{1}{r^2}](https://tex.z-dn.net/?f=A%20%5Csim%20%5Cfrac%7B1%7D%7Br%5E2%7D)
In this problem, the distance is decreased to hald the initial distance, so we can write
![r'=\frac{r}{2}](https://tex.z-dn.net/?f=r%27%3D%5Cfrac%7Br%7D%7B2%7D)
therefore, the new amplitude will be
![A' \sim \frac{1}{r'^2}=\frac{1}{(\frac{r}{2})^2}=4 \frac{1}{r^2}=4 A](https://tex.z-dn.net/?f=A%27%20%5Csim%20%5Cfrac%7B1%7D%7Br%27%5E2%7D%3D%5Cfrac%7B1%7D%7B%28%5Cfrac%7Br%7D%7B2%7D%29%5E2%7D%3D4%20%5Cfrac%7B1%7D%7Br%5E2%7D%3D4%20A)
So, the amplitude increases by a factor 4.
Answer:
W = 7.53 inches
H = 30.14 inches
Explanation:
Printing area is:
![A_p=W_o*H_o=50 in^2](https://tex.z-dn.net/?f=A_p%3DW_o%2AH_o%3D50%20in%5E2)
Solving for
:
![H_o=50/W_o](https://tex.z-dn.net/?f=H_o%3D50%2FW_o)
The total area of the paper is:
Replacing
:
![A_t=(W_o+4)*(50/W_o+16)](https://tex.z-dn.net/?f=A_t%3D%28W_o%2B4%29%2A%2850%2FW_o%2B16%29)
If we derivate and equal zero to find the minimum:
![A_t'=(50/W_o+16)+(W_o+4)*(-50/W_o^2)](https://tex.z-dn.net/?f=A_t%27%3D%2850%2FW_o%2B16%29%2B%28W_o%2B4%29%2A%28-50%2FW_o%5E2%29)
![0=(50/W_o+16)+(W_o+4)*(-50/W_o^2)](https://tex.z-dn.net/?f=0%3D%2850%2FW_o%2B16%29%2B%28W_o%2B4%29%2A%28-50%2FW_o%5E2%29)
![0=16-200/W_o^2](https://tex.z-dn.net/?f=0%3D16-200%2FW_o%5E2)
![W_o=\sqrt{200/16}](https://tex.z-dn.net/?f=W_o%3D%5Csqrt%7B200%2F16%7D)
So,
![H_o=14.14-inch](https://tex.z-dn.net/?f=H_o%3D14.14-inch)
The final dimensions of the paper are:
W = 3.53 + 4=7.53-inch
H = 14.14 + 16= 30.14-inch
Answer:
Explanation:
Let force acting on the card = F . let its x component be F_x
F_x . 3 = 300
F_x = 100
similarly ,
F_y . 4 = 400
F_y = 100
Total F = √ 3² + 4²
= 5 N
2 ) Force of the cart on the student = 5 N ( Reaction force)
3) Distance moved by the cart = 3 + 4 = 7m
4) magnitude of displacement = 5 m