Answer:
0.09kg of air
Explanation:
The dimensions of the room are given
change the height to meters by dividing it by thousand.
For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).
Volume refers to the amount of space inside a box or a object.
The amount of air is equal to the volume.
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope ![\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]](https://tex.z-dn.net/?f=%5Cfrac%7BX%7D%7BY%7D%3D%20%5B%5Cfrac%7B1%2B2%20%5Cepsilon%20%28%5Comega%2F%20W_n%29%5E2%7D%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%2B%282%20%5Cepsilon%20%20%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%7D%5D)
where;


= 
= 4.0124
replacing them into the above equation and making X the subject of the formula:



Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV

ΔU =18.79 mJ
Answer:
35 m
0.56 m/s west
Explanation:
A) Total distance is the length of the path taken.
30 m + 5 m = 35 m
B) Velocity is displacement over time. Displacement is the difference between the final position and the initial position.
If west is -x, and east is +x, then:
Δx = -30 m + 5 m
Δx = -25 m
v = Δx / t
v = -25 m / 45 s
v = -0.56 m/s
v = 0.56 m/s west