in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.
your temp should be converted in kelvin
variables:
p1=3.0×10^6 n/m^2
t1= 270k
just add 273 to your celcius
p2= ? your solving for this
t2= 315k
then you set up the equation
(3.0×10^6)/270= (x)(315)
you then cross multiply
(3.0×10^6)315=270x
distribute the 315 to the pressure.
9.45×10^8=270x then you divide 270 o both sides to get
answer
3.5×10^6 n/m^2
You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
0.04 mm Hg / mL / min .
Explanation:
Arterial pressure = 120 mm Hg
right atrial pressure = 0 mm Hg
Drop in pressure due to peripheral resistance = 120 mm Hg
volume of cardiac output per minute = 3000 mL/min
total peripheral resistance
= 120 / 3000 mm Hg / mL / min
= 0.04 mm Hg / mL / min .
