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zheka24 [161]
2 years ago
11

You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

Physics
1 answer:
Viefleur [7K]2 years ago
5 0

Explanation:

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet

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A 7 cm thick and 20 cm long wedge is used to pierce a 3cm long log of diameter 20cm what is the velocity ratio of the wedge
stiks02 [169]

Answer:

The velocity ratio of the wedge is 0.15

Explanation:

Use the following formula to calculate the velocity ratio

Velocity Ratio = Distance Moved by effort / Distance moved by Load

Where

Distance Moved by effort = 3cm

Distance moved by Load = 20 cm

Placing values in the formula

Velocity Ratio = 3cm / 20 cm

Velocity Ratio = 0.15

6 0
2 years ago
PLEASE HELP WITH ONE QUESTION!
nata0808 [166]
I think it is -3.99 x 102 j
6 0
3 years ago
A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down
KIM [24]
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
   EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
   EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
   Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
6 0
3 years ago
2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
AnnyKZ [126]

Answer:

b. negative

Explanation:

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3 0
2 years ago
Read 2 more answers
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

3 0
3 years ago
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