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Agata [3.3K]
3 years ago
13

An electric fence displays a warning sign about the voltage and amperage of the fence. How would amperage and voltage affect the

power of the fence?
Physics
1 answer:
Lorico [155]3 years ago
4 0
V = volts 
<span>I = amps </span>
<span>P = rate or energy transfer (power) </span>
<span>and </span>
<span>P = V * I</span>
You might be interested in
One point charge has a magnitude 5.4x10-7 C. A second charge 0.25 meters away has a magnitude of 1.1x10-17 C. What is the electr
choli [55]

The electric force on the charges will be equal and opposite in nature and the magnitude will be 8.5536 × 10⁻¹³ Newton.

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

In question we have given

value of q1 = 5.4 x 10⁻⁷ C

value of q2 = 1.1 x 10⁻¹⁷ C

distance between the (r) = 0.25 m

Applying the formula

F = ( K x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 10⁹ ) x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 5.4 × 1.1) × ( 10⁹ × 10⁻⁷ x 10⁻¹⁷) )/ ( 6.25 × 10⁻² )

F = ( 53.46 × 10⁻¹⁵) / ( 6.25 × 10⁻² )

F = 8.5536 × 10⁻¹³ Newton

Electrostatic force = 8.5536 × 10⁻¹³ Newton

So, The point charges are possessing equal and opposite electrostatic force of magnitude 8.5536 × 10⁻¹³ Newton.

Learn more about Electrostatic Force here:

brainly.com/question/23121845

#SPJ10

8 0
2 years ago
Birds resting on high-voltage power lines are a common sight. the copper wire on which a bird stands is 1.28 cm in diameter and
nekit [7.7K]

Answer:

8\cdot 10^{-4} V

Explanation:

First of all, let's find the cross-sectional area of the copper wire. The radius of the wire half the diameter:

r=\frac{d}{2}=\frac{1.28 cm}{2}=0.64 cm=6.4\cdot 10^{-3} m

So the area is

A=\pi r^2 = \pi (6.4\cdot 10^{-3} m)^2=1.29\cdot 10^{-4} m^2

Now we can calculate the resistance of the piece of copper wire between the bird's feet, with the formula:

R=\rho \frac{L}{A}

where

\rho=1.68\cdot 10^{-8} \Omega m is the resistivity of copper

L=4.12 cm=4.12 \cdot 10^{-2} m is the length of the piece of wire

A=1.29\cdot 10^{-4} m^2 is the cross-sectional area

Substituting, we find

R=(1.68\cdot 10^{-8} m^2)\frac{4.12\cdot 10^{-2} m}{1.29\cdot 10^{-4} m^2}=5.4\cdot 10^{-6} \Omega

And since we know the current in the wire, I=149 A, we can now find the potential difference across the body of the bird, by using Ohm's law:

V=IR=(149 A)(5.4\cdot 10^{-6} \Omega)=8\cdot 10^{-4} V

4 0
3 years ago
Which of the following decrease when an object becomes colder?
il63 [147K]
I think the answer b
6 0
3 years ago
Directions: Consider a 2-kg
nignag [31]

The definition of mechanical energy allows to find the results for the questions about the energy changes in the movement of the system are:

     1) The midpoint of the path. Em = 784 J

     2) Lowest point of the trajectory Em = 784 J

     3) Acceleration of gravity is: g = 9.8 m/s²

     4) The bowling ball has more kinetic energy as its height decreases.

<h3>Mechanical energy.</h3>

Mechanical energy is the sum of the kinetic energy due to motion and the potential energies of the body. In the case of no friction the energy remains constant at all points, this is one of the most important principles of physics.

          Em = K + U

          Em = ½ m v² + m g h

Where Em is the mechanical energy, K and U the kinetic and potential energy, respectively, m the mass, v the velocity. G the acerleacion of gravity and h the height where the body is.

In this case when the ball is on top of the building of height y = 40 m, all its energy is potential.

           Em = U = m g h

           Em = 2 9.8 40

            Em = 784 J

when the body falls it begins to acquire speed, therefore its kinetic energy increases and the height decreases, therefore its potential energy decreases, but the sum of the two remains constant.

1) Let's calculate the energy at the midpoint of the path.

They indicate that the speed is v= 19.8 m/s and the height is y2 = 20m

         Em = ½ m v² + m g h

         em = ½ 2 19.8 ² + 2 9.8 20

         em = 392 + 392

         Em = 784 J

We see that the kinetic and potential energy have the same value and the mechanical energy maintains its initial value.

2) We calculate the energy at the lowest point of the trajectory.

They indicate that the speed is v= 28 m/s and = height is zero h=0

Let's calculate

          Em = ½ m v² + m g h

          Em = ½ 2 28² + 0

          Em = 784 J

At this point, all energy is kinetic and mechanical energy is conserved.

3) The acceleration of gravity comes from Newton's second law where the force is the universal attraction.

           F = G \frac{Mm}{r^2}

           F= ma

           \G frac{Mm}{R^2}

In the case of a body near thesurfacee of the distance of the radius of the planet and the acceleration is called gravity

            g = a = G \frac{M}{R_e^2 }

            g = \frac{ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}}{6.37 \ 10^6 }

            g = 9.81  m/s²

4) The bowling ball has more kinetic energy as its height decreases, at the point where the height is half the kinetic and potential energy are equal.

In conclusion using the definition of mechanical energy we can find the results for the questions about the energy changes in the movement of the system are:

     1) The midpoint of the path. Em = 784 J

     2) Lowest point of the trajectory Em = 784 J

     3) Acceleration of gravity is: g = 9.8 m/s²

     4) The bowling ball has more kinetic energy as its height decreases.

Learn more about mechanical energy here:  brainly.com/question/24443465

7 0
3 years ago
In an electric shaver, the blade moves back and forth over a distance of 2.00 mm. The motion is simple harmonic, with frequency
Jlenok [28]

Answer:

(A)  0.001 m  

(B) 0.75 m/s  

(C) 5.7 x 10^{2} m/s^{2}      

Explanation:

distance (d) = 2 mm

frequency (f) = 120 Hz

(A) amplitude (A) = half the distance moved by the blade

      amplitude = 2 / 2 = 1 mm = 0.001 m

(B) maximum blade speed (v) = ωA = 2πfA

     v = 2 x 3.14 x 120 x 0.001 = 0.75 m/s  

(c) maximum blade acceleration (a) = w^{2}A

     a = (2 x 3.14 x 120)^{2}x0.001 = 5.7 x 10^{2} m/s^{2}

5 0
3 years ago
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