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dedylja [7]
3 years ago
8

A wire of length L is wound into a square coil with 167 turns and used in a generator that operates at 60.0 Hz and 120 V rms val

ue in a 0.041-T magnetic field. What is the length L of the wire used to construct the coil
Physics
1 answer:
Mrac [35]3 years ago
8 0

Answer:

171.43m

Explanation:

First, define emf( electromotive force )-is the unit electric charge imparted by an energy source. In this case the generator.

The peak emf is:

E_p_e_a_k=\sqrt2(E_m_a_x)=\sqrt(2\times 120V)=170V

Substituting w=2\pi f and the value for peaf E gives:

Total length=4\sqrt {\frac{NE_p_e_a_k}{Bw}=4\sqrt{\frac{167\times 170}{0.041T\times 2pi \times 60.0Hz}

=171.43m

Hence, wire's length is 171.43m

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jek_recluse [69]
Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days
6 0
3 years ago
Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow
Inessa [10]

The magnitude of the current in wire 3  is 2.4 A and in a direction pointing in the downward direction.

  • The force per unit length between two parallel thin current-carrying I_1 and I_2  wires at distance ' r ' is given by  f=\frac{u_0I_1I_2}{2\pi r}   ....(1) .
  • If the current is flowing in both wires in the same direction, and  the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

F_2_1=F_2_3

Using equation (1) , we get

\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

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5 0
1 year ago
At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s
Mumz [18]

Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

    M = \dfrac{W}{g}

    M = \dfrac{40}{9.8}

           M = 4.08 Kg

b) Mass on the surface of Lo

 Mass of an object remain same.

  Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

   W' = m g'

   W' =4.08 x 1.81

   W' = 7.38 N

8 0
2 years ago
2. A powerful experimental sewing machine is powered by a mass-spring system. This
Alexus [3.1K]

We have that the Number of stitches per sec and he mass of  oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of  oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec  is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

b)

Generally the equation for the Time t  is mathematically given as

T=2\pi\sqrt{\frac{m}{k}}

Therefore

0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}

m=2.033e-5kg

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7 0
2 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
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