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Andre45 [30]
3 years ago
15

A small circular loop of 5 mm radius is placed 1 m away from a 60-Hz power line. The voltage induced on this loop is measured at

0.6 µV. What is the current on the power line

Physics
1 answer:
marin [14]3 years ago
5 0

Answer:

i = 101.4A

Explanation:

The steps to the solution can be found in the attachment below.

We have been given the frequency f = 60Hz. From this we can calculate the angular frequency of the power line.

It is assumed that sinwt = –1 in order to calculate the time varying current. Although the magnitude of the current is large, consideration should also be given to the distance between the coil and the power line. The induced emf is small considering the area of the coil which is 7.85×10-⁵m².

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A 10 Kg dog is running with a speed of 5.0 m/s what is the minimum work required to stop the dog
ra1l [238]

Answer:

25J

Explanation:

Given parameters:

Mass of the dog  = 10kg

Speed of the dog  = 5m/s

Unknown:

The minimum energy required to stop the dog  = ?

Solution:

The dog is moving with a kinetic energy and to stop the dog, an equal amount of kinetic energy generated must be applied to the dog.

 To find the kinetic energy;

        K.E  = \frac{1}{2} m v²

m is the mass

v is the velocity

Now insert the parameters and solve;

      K.E  =  \frac{1}{2}  x 10 x 5  = 25J

8 0
3 years ago
A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th
jok3333 [9.3K]
<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

              s = ut + 0.5 at²

              1.22 = 10.36 x t + 0.5 x -9.81 x t²

              4.905t² - 10.36 t + 1.22 = 0

              t = 1.99 s     or    t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

6 0
3 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
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