Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:
Differentiating with respect to time we get
a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0
b) The object will reach the highest point after 20 seconds
c) Highest point the object will reach is 360 m
d) Time taken to strike the ground would be 20+20 = 40 seconds
![[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s](https://tex.z-dn.net/?f=%5Btex%5Dv%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B0.9%5Ctimes%202%5Ctimes%2020%5C%5C%5CRightarrow%20v%3D36%5C%20m%2Fs)
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of 
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³
Answer:
D.Entropy tends to increase.
Explanation:
The second of thermodynamics states that the state of entropy of the entire universe,as an isolate system, will always increase over time
<span>The person is dragging
with a force of 58 lbs at an angle of 27 degrees relating to the ground. We
want to use cosine function to look for the horizontal force component. And
then we can compute for W = (Horizontal Force) x (Distance). We want the
horizontal force component since that is the component that is parallel to the
direction the cart is moving. </span><span>
(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal
force component.)
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>
Answer:
Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.
Explanation:
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