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Genrish500 [490]
3 years ago
8

What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?

Physics
1 answer:
eimsori [14]3 years ago
4 0

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

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3 years ago
A circular loop of wire of area 10 cm^2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is
S_A_V [24]

The magnetic dipole moment of the current loop is  0.025 Am².

The magnetic torque on the loop is 2.5 x 10⁻⁴  Nm.

<h3>What is magnetic dipole moment?</h3>

The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.

Mathematically, magnetic dipole moment is given as;

μ = NIA

where;

  • N is number of turns of the loop
  • A is the area of the loop
  • I is the current flowing in the loop

μ = (1) x (25 A) x (0.001 m²)

μ = 0.025 Am²

The magnetic torque on the loop is calculated as follows;

τ = μB

where;

  • B is magnetic field strength

B = √(0.002² + 0.006² + 0.008²)

B = 0.01 T

τ = μB

τ =  0.025 Am² x 0.01 T

τ = 2.5 x 10⁻⁴  Nm

Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.

Learn more about magnetic dipole moment here: brainly.com/question/13068184

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8 0
1 year ago
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various
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Answer:

q  =  -461532.5 \ C

Explanation:

From the question we are told that

     The  electric filed is  E  =  102 \ N/C  

Generally according to Gauss law

=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

          A is the surface  area of the of the earth which is mathematically represented as

     A  =  4\pi r^2

Where r is the radius of the earth which has a value r = 6.3781*10^6 m

 substituting values

    A  = 4 * 3.142  *   (6.3781*10^6 \ m)^2

    A  =5.1128 *10^{14} \ m^2

So

   q  =  -E  * A  *  \epsilon _o

Here \epsilon_o s the permitivity of free space with value

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substituting values

     q  =  -102  * 5.1128 *10^{14}  *  8.85 *10^{-12}

     q  =  -461532.5 \ C

6 0
3 years ago
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Answer:

<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>

<h2>FORMULA OF PRESSURE = </h2>

<h2>P=F/A </h2>

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3 years ago
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