Answer:
Using the new cylinder the heat rate between the reservoirs would be 50 W
Explanation:
- Conduction could be described by the Law of Fourierin the form:
where
is the rate of heat transferred by conduction,
is the thermal conductivity of the material,
and
are the temperatures of each heat deposit,
is the cross area to the flow of heat, and
is the distance that the flow of heat has to go. - For the original cylinder the Fourier's law would be:
, and if
, then the expression would be:
where
is the diameter of the original cylinder, and
is the length of the original cylinder. - For the new cylinder, in the same fashion that for the first, Fourier's Law would be:
,where
is the heat rate in the second case,
and
are the new diameter and length. - But,
and
, substituting in the expression for
:
. - Rearranging:
. - In the last declaration of
, it could be noted that the expressión inside the parenthesis is actually
, then:
. - <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg
choose A
A. 4 cm behind the mirror
<span> For any mirror, </span><span><span>so</span><span>si</span>=<span>f^2</span></span>. Therefore, by plugging in the values, you get <span>18<span>si</span>=144. 144/18 = 8, so </span><span><span>si</span>=8</span>
<span> The focal point is located 12 cm from the mirror, and </span><span>si</span><span> is the distance of the image from the focal point, so the image is 4 cm from the mirror. The mirror is convex, so then the focal point and the image are both behind the mirror.</span>
Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa