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Anton [14]
3 years ago
5

Which of the following statements most accurately describes the circuit?

Engineering
1 answer:
Fofino [41]3 years ago
7 0

Answer:

D

Explanation:

All of the above.

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Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregar
suter [353]
Search up A gardener can increase the number of dahlia plants in an annual garden by either buying new bulbs each year or dividing the existing bulbs to create new plants . The table below shows the expected number of bulbs for each method

Part A
For each method,a function to model the expected number of plants for each year

Part B
Use the Functions to Find the expected number of plants in 10 years for each method.

Part C
3 0
3 years ago
It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore
Dmitry [639]

Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

C. 9.1125 x 10^14 kg of CO2

Explanation:

A. Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Initial area of the rainforest = 20 x 10^12 m^2

Therefore to calculate exponential rate in 1/day,

Rate of rainforest destruction/ initial area of rainforest

= 139.2 x 10^6/20 x 10^12

= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

(40 * 365 )days + 10 days (leap years)

= 14610 days

Area of rainforest in 1975 = 24.5 x 10^12m^2

Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

= 2.034 x 10^12 m^2

Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

= 24.5 x 10^12 - 2.034 x 10^12

= 22.466 x 10^12 m^2

C. How much CO2 will be removed in 2025

Recall: Photosynthesis is the process of plants taking in CO2 and water to give glucose and O2.

So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

6 0
3 years ago
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
3 years ago
The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
NemiM [27]

Answer:

More Drag on the down going wing and More Lift on the up going wing

Explanation:

The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.

Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.

As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.

7 0
3 years ago
What steps would you take to design an improved toothpaste container?
algol [13]

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0
3 years ago
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