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Lapatulllka [165]
3 years ago
9

how can you find the mechanical advantage of the six different simple machines? (simples mechinals are pulleys, inclined plane,

wheel and axle, lever, wedge, and screw)
Physics
1 answer:
abruzzese [7]3 years ago
7 0

Hey there!

Each machine preety much have the same type of concept. Let's take for example, a screw. What is a screw? A screw is simply a long/small little "stick" that is able to 'twurl' it's way inside a hole, and hold it's self tight, and also take it's self right back out.

Now, based on the other 5 machines, we can see that it would be very similar of the screw. There would be a in and out function, left to right function, up and down function, and a rounded function, such as a pulley. Those would be the advantages of these different types of machines.

I hope this helps you!

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Brrunno [24]

Answer:

Covalent Bond is found between the atoms of a molecule.

4 0
3 years ago
Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc
Ira Lisetskai [31]

Answer:

0 < r < r_exterior     B_total = \frac{\mu_o I}{2\pi  r}

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = \frac{\mu_o I}{2\pi  r}

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = \frac{\mu_o I}{2\pi  r}  

r > r_exterior

          B_total = B₁ -B₂

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6 0
3 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that incre
olga nikolaevna [1]

Answer:

The  induced current is I  =  6.25*10^{-4} \  A

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    The number of turns is  N  =  1

     The  cross-sectional area is  A  =  8.20 cm^2  =  8.20 * 10^{-4} \ m^2

    The  initial magnetic field is  B_i  =  0.500 \ T

     The  magnetic field at time =  1.02 s  is  B_t =  2.60 \ T

     The  resistance is  R  = 2.70\  \Omega

The  induced emf is mathematically represented as

       \epsilon  = - N  *   \frac{ d\phi }{dt}

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

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        d \phi  =  dB  *  A

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        dB  =  B_t  - B_i

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     |\epsilon|  = 1.69 *10^{-3} \  V

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      I  =  \frac{\epsilon}{ R }

  substituting values

       I  =  \frac{1.69*10^{-3}}{ 2.70 }

       I  =  6.25*10^{-4} \  A

7 0
4 years ago
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