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Marrrta [24]
3 years ago
9

What does Rockstrom mean by the need to "bend the curves"?

Physics
1 answer:
umka21 [38]3 years ago
4 0

Answer: Rockstrom meant that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.

Explanation:

Some years ago, Johan Rockström, who was the executive director of Stockholm Environment Institute, and a professor at Stockholm University, was the head of an international team that was assembled in defining planetary boundaries.

The that was gathered to speak regarding issues that pertained to the Earth ane how it can be protected from being infiltrated ad protecting it from failing. During the conversation, Rockstrom said that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.

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Which observations correctly describe the second law of thermodynamics? Check all that apply.
melomori [17]

Answer:

Thermal energy flows from a warmer object to a cooler object.

Explanation:

The second law of thermodynamics speaks about the direction of processes. Clarifying that these can only go in one direction as time progresses in one direction. To better understand this concept, several examples will be explained.

The first example is when we have a wood that we will use to make a campfire, as we require thermal energy to heat a space. We see that as the fire consumes the wood heat is generated inside the space. This will happen until the wood is consumed. Now if we want to perform the reverse process, that is to cool the room and use the heat content in this to generate wood, we see that it is an impossible process to perform, that is to say, it is true that this thermal process has only one direction.

Now suppose we have a cup of coffee at a temperature of 95 (°C), with the passage of time the coffee cools this is because it generates a heat transfer from the coffee to the surroundings. Similarly if we want to collect the heat from the surroundings and pass them to the coffee so that it reaches a temperature close to 95 (°C), naturally it can not be done, the heat transfer always exists in one direction this is from objects at high temperature to objects of lower temperature.

To violate the second law of thermodynamics, and to allow processes to go in the opposite direction, it is only possible through mechanical work that adheres to the process. For example if we have a canned drink at 25 (°C) and want to reduce the temperature of this drink, then we must put it in a refrigerator, so that it can reduce its temperature to 5 (°C). The refrigerator is electromechanical equipment that can alter heat transfer directions.

3 0
4 years ago
The velocity profile in a water tunnel was measured on the upstream and downstream sides of an object in order estimate the drag
alexdok [17]

Answer:

R_x = 49.78 lb/ft

Explanation:

Given:

- The velocity u_1 upstream = 10 ft/s

- Pressure field p_1 = p_2 = 10 psig

- The velocity profile downstream is given by:

                       u_2 = 10 ft/s         ...... |y| >= 2

                       u_2 = 10 - 3/8*(2 - |y| )^2   ....... |y| < 2

- The profiles are given in attachment.

Find:

Determine the drag force per unit length.

Solution:

- Develop a control volume around the object see attachment which only consists of water. Then apply principle of linear momentum along the x-direction. It can be expressed as:

                       -p_w*u_1^2*A_1 + 2 \int\limits^2_0 {u_2^2*p_w} \, dy = -F_x\\p_w*u_1^2*h - 2*p_w \int\limits^2_0 {u_2^2} \, dy = R_x

Where, p_w: Density of water = 1.94 slugs/ft^3

            h : Vertical height of control volume

            R_x: The reaction force exerted by object on control volume(Drag)

- To determine h, The conservation of mass principle is applied at sections 1 and 2:

                         p_w*u_1*h = 2\int\limits^2_0 ({p_w*u_2}) \, dy \\h = \frac{2}{u_1} \int\limits^2_0 ({10-\frac{3*(2-y)^2}{8} } )\, dy \\\\h = \frac{2}{u_1} \limits^2_0 ({10y+\frac{(2-y)^3}{8} } )\,\\\\h = \frac{2}{10}*({20 + 0 - 0-1 ) = 3.8 ft

- Now for Drag force per unit length we have:

                         p_w*u_1^2*h - 2*p_w \int\limits^2_0 {(10-\frac{3*(2-y)^2}{8} )^2} \, dy = R_x\\\\p_w*u_1^2*h - 2*p_w \int\limits^2_0 ({100+\frac{9*(2-y)^4}{64}-\frac{15*(2-y)^2}{2}}) \, dy = R_x\\\\p_w*u_1^2*h - 2*p_w * ({100y-\frac{9*(2-y)^5}{320}+\frac{5*(2-y)^3}{2}}) \limits^2_0 = R_x\\\\1.94*10^2*3.8 - 2*1.9*({200+\frac{9}{10}-20}) = R_x\\\\R_x = 737.2 - 687.42\\\\R_x = 49.78 lb/ft        

- The drag force per unit length on the object is given by R_x = 49.78 lb/ft. It is also the reaction developed due to change in momentum of fluid.

3 0
3 years ago
A car drives horizontally off the edge of a cliff that is 40.8 m high. the police at the scene of the accident note that the poi
Inga [223]
Fall time machine:
t = √ (2*H/g) = √ (2*40.8 / 9,8) ≈ 2.9 s
Horizontal movement:
S = V₀*t
Speed:
V₀ = S / t = 121 / 2.9 ≈ 41.7 m/s         or  150 km/h

7 0
3 years ago
You are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You p
gladu [14]

Answer:

v = 133.5 km/h

Explanation:

As we know that the distance moved by the car is given 300 km

first it move for 120 km with speed

v = 100 km/h

so the time taken by car to move this distance is

t = \frac{d}{v}

t = \frac{120}{100}

t = 1.2 h

Now it moves next 43.0 km with speed 42 km/h

so the time taken by it is given as

t = \frac{43}{42}

t = 1.02 h

now total time of journey is

t = 11:15 AM - 8 AM = 3.25 h

time remaining to cover the last part of journey is

\Delta t = 3.25 - (1.02 + 1.2) = 1.026 h

now the remaining distance is

d = 300 - 120 - 43 = 137 km

so the speed of the car is given as

v = \frac{137}{1.026}

v = 133.5 km/h

8 0
3 years ago
I WILL NEVER USE THIS APPP AGAIN I JUST GOT A 63 PPERCENT OFF PEOPLES ANSWERS AND GOT A 63% WHAT IS THAT
geniusboy [140]

Answer:

You should still use the app if you need it, but you definitely shouldn't rely on it. It's really good for what it is, but it's not an alternative to actually taking the test, just imagine what you would have gotten if you didn't use this

Explanation:

4 0
3 years ago
Read 2 more answers
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