People in a community are interested in constructing a hydroelectric dam in
order to get more affordable electricity. Which option is best described as an
engineering endeavor related to building a hydroelectric dam?
O A. Taking measurements and performing calculations to determine
how strong the dam needs to be
B. Making observations in order to determine what kinds of plants
grow near the proposed dam site
C. Studying the rocks around the proposed dam site to determine
how old they are
D. Performing an investigation to determine how the construction will
affect local ecosystems
SUBMIT

Answers

Answer 1

People  in a community are interested in constructing a hydroelectric dam in order to get more affordable electricity, so the next best option is to take measurements and perform calculations to determine how strong the dam needs to be, that is option A.

What is the work of an engineer?

For building a hydroelectric dam, the engineers must take accurate measurements and perform detailed calculations so that the strength and stability of the dam are maintained, so the work of the engineers is important in construction.

Hence, people in a community are interested in constructing a hydroelectric dam in order to get more affordable electricity, so the next best option is to take measurements and perform calculations to determine how strong the dam needs to be, that is option A.

Learn more about the engineer here.

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Related Questions

A car company is doing crash tests and has installed forcesensors in the chairs and seatbelts to measure the forces thatare exerted on crash test dummies during a crash. In oneparticular test the sensors measure a horizontal net force onthe dummy of 39,400 N when the car comes to a complete stopfrom an initial speed of 35 m/s in 1.4 m. Treating the dummyas a point particle and assuming it comes to rest in the samedistance, what is the mass of the test dummy used during thistest

Answers

Answer:

[tex]m=90.1kg[/tex]

Explanation:

From the question we are told that

Force F=39400

Initial speed [tex]V_1=35m/s[/tex]

Distance[tex]d=1.4m[/tex]

Generally the equation for acceleration is mathematically given by

[tex]a=\frac{V^2-U^2}{2d}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=-437.5m/sec^2[/tex]

Generally the equation for mass is mathematically given by

[tex]F=ma\\m=F/a[/tex]

[tex]m=\frac{-39400}{-437.5}[/tex]

[tex]m=90.1kg[/tex]

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answers

This question is incomplete, the complete question is;

A positive charge of magnitude Q1 = 6.5 nC is located at the origin.

A negative charge Q2 = -3.5 nC is located on the positive x-axis at x = 16.5 cm from the origin. The point P is located y = 10.5 cm above charge Q2.

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answer:

the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C    

Explanation:

Given the data in the question;

Q1 = 6.5 nC, Q2 = -3.5 nC

from the image below, to get our angle ∅

tan∅ = opp/adj

tan∅ = 10.5 / 16.5

tan∅ = 0.636363

∅ = tan⁻¹( 0.636363 )

∅ = 32.47°

also, r1 = √( 16.5² + 10.5²)

r1 = √( 272.25 + 110.25 )

r1 = √382.5

r1 = 19.55 cm = 0.1955 m

Now, the x-component of the electric field at point P due to charge Q1 will be;

Ex = E2cos32.47°

= (kQ1/r1²)cos32.47°

we know that;  k is Coulomb's law constant ( 9 × 10⁹ N.m²/ C²

Q1 = 6.5 nC = 6.5 × 10⁻⁹ C

so we substitute

= ((9 × 10⁹ × 6.5 × 10⁻⁹) / (0.1955)²) cos32.47°

= (58.5 / 0.03822025) × 0.843672

= 1291.33 N/C

Therefore, the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C

during what part of their drive are the not moving?

Answers

C answer is c math ez
The correct answer is c

A 800-w iron is left on the ironing board with the base exposed to the air. About 85 percent of the heat generated by iron dissipate through its base with surface area of 0.15 ft2 . Calculate the amount of heat the iron dissipates in 2 hours, the heat flux on the surface of iron and total cost of electricity wasted in 2 hours if unit cost of electricity to be $0.11kwh.

Answers

Answer:

ΔQ = 4896 KJ

Heat Flux = 57407.55 W/m²

Cost = $ 0.176

Explanation:

The amount of heat dissipated in two hours can be given as:

[tex]Heat\ Dissipation = \Delta Q = Pt\eta[/tex]

where,

P = Power of Iron = 800 W

t = time taken = 2 hrs

η = Conversion efficiency = 85% = 0.85

[tex]\Delta Q = (800\ W)(7200\ s)(0.85)\\\\[/tex]

ΔQ = 4896 KJ

Now, for heat flux:

[tex]Heat\ Flux = \frac{P}{Surface\ Area}\\\\Heat\ Flux = \frac{800\ W}{0.15\ ft^{2}}\frac{1\ ft^{2}}{0.092903\ m^{2}}[/tex]

Heat Flux = 57407.55 W/m²

Now, for total heat energy in KWhr in two hours:

[tex]Heat\ Energy = (Power)(Time)\\\\Heat\ Energy = (0.8\ KW)(2\ hr)\\\\Heat\ Energy = 1.6\ KWh[/tex]

Now, for the cost:

[tex]Cost = (Heat\ Energy)(Unit\ Cost)\\Cost = (1.6\ KWh)(\$ 0.11\ /KWh)[/tex]

Cost = $ 0.176

Can anyone help me with the Wet Lab Guide - Coulomb's Law report? I'm really having trouble with it. I have attached the worksheet

Answers

Answer:

Coulomb's law of Electrostatic Force states that a charged particle attracts or repels other charged particles with a force which is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers.

Explanation:

can someone help me :(​

Answers

Answer:

1. This is a speed graph.

2. The slope represents how much the item/ thing is moving at how much distance it moved and the minutes it takes it too move how much distance, but the slope is representing the decrease and increase of it.

3. Increasing, kind of rapidly going up since the slope is rising not falling.

Besides gravity, what factor keeps the moon and Earth in orbit?

force
inertia
mass
distance

Answers

Answer:

interna

Explanation:

please mark as brainllest

Answer is inertia
Good Luck ;)

If a large number of waves pass by a certain point in a given time, the waves have a low frequency. none of the above. a high frequency. it doesn't affect frequency.

Answers

Answer:

Another term used to describe a wave is frequency. Since waves are moving, we define frequency as the number of waves that pass a given point in a specified unit of time. The unit commonly used is Hertz which is the number of wave cycles pass a point in one second. So one cycle per second equals one Hertz

To start a great night of doing physics homework, you sit down to pour yourself a good cup of coffee. Your coffee mug has a mass 137 g and a specific heat of 1089 J/kg K. The mug starts out at room temperature (23.8 o C). Your coffee has an initial temperature of 79.8 oC and has the same specific heat as water (4186 J/kg K).
1) Let's say you pour enough coffee into the mug so that the mass of coffee is equal to the mass of the mug. If we assume that no heat is lost to the outside room, once the mug and coffee reach the same temperature, will that temperature be closer to the initial temperature of the coffee or the initial temperature of the mug? (Notice that the mass of the coffee and the mass of the mug are the same.)
2) What is the final temperature of coffee and mug once they come to thermal equilibrium?
Tfinal = ____________ degrees c
3) Now lets assume that instead of the 137 g of coffee, you pour in 225 g of coffee. What is the final temperature of the coffee and mug? (Again, assume that you loose no heat to the outside.)
Tfinal = ____________ degrees c
4) Now lets say that along with the 225 g of coffee, you pour in 11.7 g of cream in your mug. The cream has an initial temperature of 5.2 oC and also has the same specific heat as water. What is the final temperature of the coffee, cream and mug? (Again, assume that you loose no heat to the outside.)
Tfinal = _____________ degrees c

Answers

Answer:

1) The temperature will be closer to water

2) T = 68.239°C

3) T = 72.142°C

4) T = 69.266 °C

Explanation:

1)

The temperature will be closer to water because  the heat capacity of water > heat capacity of coffee.

2)

137(1.089)(T - 23.8) = 137(4.186)(79.8 - T)

⇒(1.089)(T - 23.8) = (4.186)(79.8 - T)

⇒1.089 T - 25.9182 = 334.0428 - 4.186 T

⇒1.089 T +  4.186 T = 334.0428 + 25.9182

⇒5.275 T = 359.961

⇒ T = 68.239°C

3)

137(1.089)(T - 23.8) = 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 = 75159.63 - 941.85 T

⇒149.193 T +  941.85 T = 75159.63  + 3550.7934

⇒1091.043 T = 78710.4234

⇒ T = 72.142°C

4)

137(1.089)(T - 23.8) + 11.7(4.186)(T - 5.2)= 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) + 48.9762(T - 5.2) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934  + 48.9762 T - 254.67624= 75159.63 - 941.85 T

⇒149.193 T +  941.85 T + 48.9762 T = 75159.63  + 3550.7934 + 254.67624

⇒1091.043 T + 48.9762 T = 78710.4234 + 254.67624

⇒1140.0192 T = 78965.09964

⇒ T = 69.266 °C

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