Answer:
B) 16.67
Explanation:
If the dimension of one lumber is 2" × 6", the total area of one lumber will be 12inch²
If the total board feet of lumber there is 200in, therefore the total board of lumber that will be needed is 200/12 which gives 16.67 lumbers
They can fight the infection but not the disease
Answer:
(a) 
(b) 
Explanation:
<u>Given:</u>
= The first temperature of air inside the tire = 
= The second temperature of air inside the tire = 
= The third temperature of air inside the tire = 
= The first volume of air inside the tire
= The second volume of air inside the tire = 
= The third volume of air inside the tire = 
= The first pressure of air inside the tire = 
<u>Assume:</u>
= The second pressure of air inside the tire
= The third pressure of air inside the tire- n = number of moles of air
Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.
Using ideal gas equation, we have

Part (a):
Using the above equation for this part of compression in the air, we have

Hence, the pressure in the tire after the compression is
.
Part (b):
Again using the equation for this part for the air, we have

Hence, the pressure in the tire after the car i driven at high speed is
.
The total energy equation would be Kinetic energy+Potential energy
Answer:
24cm/s
Explanation:
A=L*w
A'=L'*w'
L=13
w=5
L'=4
w'=6
A=?
A'=?
A=L*w
A=13*5
A=65
A'=L'*w'
A'=4*6
A'=24
*the given lengths are just to throw you off*