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Ganezh [65]
3 years ago
5

Please help, thank you !

Physics
1 answer:
kherson [118]3 years ago
6 0

Answer:

0.04658 m from the nut

Explanation:

Recall that torque is defined as the vector product of force (F) times distance (d). Then, given that the force is applied perpendicular to the wrench, gives the sine of the angle (90 degrees) equal to 1, simplifying the calculation which consists basically on solving for d in the equation:

2.25 N*m = 48.3 N * d

d = 2.25 /48.3  m ≈ 0.04658 m

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S= 1/2(V f + V I )t solve for V f <br><br> show me how
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<span>S= 1/2(V f + V I )t solve for V f 
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4 years ago
A rod (length = 2.0 m) is uniformly charged and has a total charge of 5.0 nC. What is the electric potential (relative to zero a
Ksenya-84 [330]

Answer:

The electric potential is  V  =  15.6 V

Explanation:

From the question we are told that

   The length  of the rod is  L =  2.0 \ m

     The total charge of the rod is  q =5.0 nC = 5.0*10^{-9} C

      The length from the center is  d =  3.0 \ m

The diagram illustrating the setup for this question is shown on the first uploaded image

From the diagram the potential at point  A  dl is mathematically represented as

         dV  =  K  \frac{dq}{l}

Where K is the coulomb constant with a value  K  = 9*10^9 \   Nm^2 /C^2

where q is the charge in charge  the rod relative to its distance from A  is mathematically represented as

         dq =  \frac{q}{L}  dl

This a small unit length of the rod

So         V = \frac{q}{L}  \int\limits^4_2  {\frac{dl}{l} } \,

        =>   V =  k\frac{q}{L}  ln [\frac{4}{2} ]

              V =  k\frac{q}{L}  ln2

Substituting values

                V  =  9* 10^9  *  \frac{5*10^{-9}}{2} * ln 2

                 V  =  15.6 V

         

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4 years ago
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The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

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Learn more about elastic potential energy: brainly.com/question/156316

4 0
2 years ago
What is the magnetic force on a proton that is moving at 4.5 x 10^7 m/s to the right through a magnetic field that is 1.6 T and
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The magnetic force on a charged particle is given as: F = qvBsinФ
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⇒ F = qvb sin Ф

Here, Ф = 90 degree.
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Putting the values in the equation, we get,

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= 11.52 × 10^-12N
= 1.2 x 10^-11 N up
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3 years ago
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