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bazaltina [42]
3 years ago
6

An accelerating voltage

Physics
1 answer:
ki77a [65]3 years ago
8 0
Accelerating voltage<span> is the difference in potential between the filament and the anode,and it can be varied between 5 KeV and 30 KeV on the S-500 and between 2 KeV and 30 KeV on the S-450. As the </span>voltage<span> is increased, the electrons travel with higher velocity and are more energetic.</span>
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When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec
igomit [66]

Answer:

The maximum electric power output is P_{max} =1.339*10^{9} \ W

Explanation:

From the question we are told that

        The capacity of the hydroelectric plant is \frac{V}{t}   =  690 \ m^3 /s

         The level at which water is been released is h  =  220 \ m

        The efficiency is  \eta  =0.90

       

The electric power output is mathematically represented as

       P  = \frac{PE_l - PE _o}{t}

Where  PE_l is the potential energy at  level h which is mathematically evaluated as  

          PE_l  =  mgh

and  PE_o  is  the potential energy at ground level which is mathematically evaluated as  

          PE_o  =  mg(0)

         PE_o  =  0

So  

         P  = \frac{mgh}{t}

here  m  =   V *  \rho

where V is volume  and  \rho is density of water whose value is  \rho = 1000 kg/m^3

 So  

         P  = \frac{(\rho * V) * gh}{t}

        P  = \frac{V}{t} * gh \rho

substituting values  

       P  =690 * 9.8 * 220 * 1000

      P  =1.488*10^{9} \ W

The maximum possible electric power output is

           P_{max} = P * \eta

substituting values  

         P_{max} =1.488*10^{9} * 0.90

         P_{max} =1.339*10^{9} \ W

6 0
3 years ago
The efficiency of a machine is always less than 100% because part of the energy input is used to
Travka [436]

Answer:

C. Overcome Friction

Explanation:

When using any machine usually those with moving parts, you may notice heat forming near the areas where most movement occurs. As friction continues, more energy is used up and released as heat. For that reason, the efficiency of a machine will forever be less than 100%

5 0
3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,
borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

5 0
3 years ago
Q= Which one of the following statement is incorrect?
Lunna [17]

Answer:

A.

Explanation:

I think it might be the big number A

8 0
3 years ago
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