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Zielflug [23.3K]
3 years ago
6

All of the following statements are true of hydroelectric power EXCEPT: A. Hydroelectric power stations rely on dams to help har

ness the energy of moving water. B. The water within hydroelectric power stations will transfer energy from gravitational potential energy to kinetic energy. C. Hydroelectric power stations can only produce enough energy for a small town as they do not produce large quantities of energy. D. Hydroelectric power stations are very reliable source of energy.
Physics
2 answers:
Artist 52 [7]3 years ago
5 0
The correct answer should be C. Hydroelectric power stations can only produce enough energy for a small town as they do not produce large quantities

Hydroelectric power stations can power even large cities that have millions of people.
kicyunya [14]3 years ago
3 0
In my opinion, the correct answer among the choices listed above is option C. <span>Hydroelectric power stations cannot only produce enough energy for a small town as they do produce large quantities of energy for a large area. Hope this answers the question.</span>
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What causes an object to change its motion?
mr_godi [17]

Answer:

An unbalanced force.

Explanation:

Teacher: "An object in motion will stay in motion unless..."

Auggie : "Acted on by an unbalanced force."

Teacher: "Very good."

3 0
3 years ago
A cargo elevator on Earth (where g = 10 m/s2) lifts 3000 kg upwards by 20 m. 720 kJ of electrical energy is used up in the proce
MakcuM [25]

Answer: 83%

Explanation:

Efficiency of the process = work output/work input × 100%

Work input is the energy used up in the process = 720,000Joules

Work output = Force × distance

= (3000×10)× 20

= 600000 Joules

Efficiency= 600000/720000 × 100

= 0.83×100

= 83%

8 0
3 years ago
A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh
otez555 [7]

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

From the question we are told that

   The center of gravity of the plane to its nose  is  z = 2.58 m

    The distance of the front wheel of the plane to  its nose is l = 0.800\ m

     The distance of the main wheel of the plane to its nose is e =  3.02 \ m

At equilibrium  the Torque about the nose of the airplane is mathematically represented as

          mg (z- l) -  G_B *(e - l) = 0

Where m is the mass of the airplane

          G_B is the weight of the airplane supported by the main wheel  

       So  

             G_B =\frac{mg (z-l)}{(e - l)}

Substituting values

            G_B =\frac{mg (2.58 -0.8 )}{(3.02  - 0.80)}

           G_B = 0.8018 mg

Now the weight supported at the frontal wheel is mathematically evaluated as

           G_F = mg - G_B

Substituting values      

       G_F = mg - 0.8018mg    

      G_F = (1 - 0.8018) mg      

     G_F = 0.1982 mg    

Now the weight of the airplane is  =  mg

Thus percentage of this weight supported by the front wheel is  A = 0. 1982 *100 = 19.82 %

7 0
3 years ago
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

5 0
3 years ago
3. What are the two types of mixtures?
ioda

Answer:

heterogeneous and homogeneous

Explanation:

Hetero you can separate whilst homo is combined and cant be separated

7 0
3 years ago
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