frequency=4Hz
wavelength=5m
amplitude=1/2×2=1m
period=1/frequency
1/4=0.25seconds.
velocity=wavelength×frequency
=5×4
=20m/s
Answer:
A. 4.82 cm
B. 24.66 cm
Explanation:
The depth of water = 19.6 cm
Distance of fish = 6.40 cm
Index of refraction of water = 1.33
(A). Now use the below formula to compute the apparent depth.
(B). the depth of the fish in the mirror.
Now find the depth of reflection of the fish in the bottom of the tank.
Answer:
Both technicians A and B are correct, the information on how to fill the car tires can be located on the tire and on the placard on the driver's door jam
Explanation:
The instruction on proper inflation of car tires can be found on the tire information placard in the door jam area or the fuel filler door, in the glove box or in the car or tire manual
The tire information of the proper pressure to fill the tires is also visibly engraved on the sides of the car tire, where other information such as the requirement for training on tire filling/changing, safety precautions and manufacturer's information
This question is in two parts. This is not the correct multiple choice options for this part a.
The second part had the option
b)If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?
a)30 m^2
b) 31.4 m^2
c)31.37 m^2
d)31.371 m^2
Answer:
A. 17.0 m²
B. 31.4 m²
Explanation:
The formula for the calculation of the area of a rectangle is given as
Area = length x width
The length = 3.547 m
The width = 4.79 m
Then area = 3.547 x 4.79
= 16.990m²
When approximated = 17.0m²
This is the most precise measurement for the area of the bedroom.
B.
We solve b using this formula
Area = pi(diameter/2)^2
= 3.14(6.32/2)²
= 3.14 x 9.9856
= 31.4 m²
Answer:
the speed of the car at the top of the vertical loop 
the magnitude of the normal force acting on the car at the top of the vertical loop 
Explanation:
Using the law of conservation of energy ;
The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:
![F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\](https://tex.z-dn.net/?f=F_%7BN%7D%20%3D%20%5Cfrac%7Bmv%5E2_%7Btop%7D%7D%7BR%7D%20%5C%20-%20mg%5C%5C%5C%5CF_%7BN%7D%20%3D%20%5Cfrac%7Bm%282.0%20%5Csqrt%7BgR%7D%29%5E2%7D%7BR%7D%20%5C%20-%20mg%5C%5C%5C%5CF_%7BN%7D%20%3D%20%5B%282.0%5E2-1%5Dmg%5C%5C%5C%5CF_%7BN%7D%20%3D%20%5B%282.0%29%5E2%20-1%29%20%2850%2A10%5E%7B-3%7D%20%5C%20kg%29%289.8%20%5C%20m%2Fs%5E2%5D%5C%5C%5C%5C)
