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Ad libitum [116K]
2 years ago
10

If the Force exerted by the intern is doubled and the distance is halved, does the done by the intern increase, decrease, or rem

ain the same?
Physics
1 answer:
Jlenok [28]2 years ago
3 0

Remain the same

Explanation:

If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.

 Work done is the force applied to move a body through a distance.

Work done = F x d

where F is the applied force

            d is the distance moved

Now;

if:

    f = 2f

    d = \frac{1}{2}d

Input the parameter:

 Work done = fxd = 2f x \frac{1}{2}d  = fd

The work done will still remain the same

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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The voltage across the primary winding is 6000 V. If the primary winding has 50 coils and the secondary winding has 20 coils, wh
sesenic [268]

Answer: 2400V

Explanation:

This is a question on transformer. A transformer is a device used to step up or step down current and voltage. It consists of the primary windings and secondary windings. Each windings possesses both coil and amount of voltage across each coil.

The transformer formula is

Vs/Vp = Ns/Np

Where;

Vs is the voltage across the secondary coil

Vp is the voltage across the primary coil

Ns is the number of turn in the primary coil

Np is the number of turns in the primary coil.

According to the question,

Vs is unknown

Vp is 6000V

Ns is 20turns

Np is 50turns

Substituting the values in the formula given

Vs/6000 = 20/50

Vs = 6000×20/50

Vs = 2400V

Therefore the secondary voltage is 2400V

4 0
2 years ago
Read 2 more answers
11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is finge
Anna35 [415]

Answer:

The frequency is  f_n  = 257.1 \ Hz

 

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  f_3 = 540 \ Hz

     

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         f =  \frac{v_s}{ 2L}

Now when it finger at 70% it original length is

      f_n  =  \frac{v}{2 *  (0.7 L)}

      f_n  =  \frac{v}{1.4 L}

Here v  the velocity of sound

  So  

         \frac{f_n}{f}  =  \frac{\frac{v}{1.4L} }{\frac{v}{2L} }

Also the fundamental frequency for the original length can also be represented as

       f =  \frac{f_3}{3}

substituting values

          f =  \frac{540}{3}

          f = 180 \ Hz

So

       \frac{f_n}{180}  =  \frac{\frac{v}{1.4L} }{\frac{v}{2L} }

=>  f_n  =\frac{180}{0.7}

=>   f_n  = 257.1 \ Hz

 

     

3 0
2 years ago
A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro
Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

                        =0.8rad/s * 5m

                        = 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

4 0
2 years ago
Kevin Tan's Balance Sheet. Total assets are 13,200 dollars. Total liabilities are 9,150 dollars. What is Kevin’s net worth on Ma
Alexus [3.1K]

Answer:

4,050

Explanation:

I did the math.

5 0
2 years ago
Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi
Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given asrate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}

(T_{1/2})_{fission} = 3*10^{17} y

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission = \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}

rate of fission =5.89*10^3 1\Year

5 0
2 years ago
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