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Dmitriy789 [7]
3 years ago
8

Match each method of transferring electric charge with the correct description

Physics
2 answers:
lions [1.4K]3 years ago
8 0

Answer:

look at the attachment for correct answers.

Lynna [10]3 years ago
4 0

Answer:

transfer of electric charge by rubbing conduction. mark me as brainest

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A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire
vitfil [10]

Answer:

a

  When r \ge R

      B =  \frac{ \mu_o *  I}{ 2 \pi r }

b

 When r< R

   B =  [\frac{\mu_o *  I }{ 2 \pi R^2} ]* r

Explanation:

From the question we are told that

   The  radius is  R  

   The  current is  I

    The  distance from the center

Ampere's law is mathematically represented as

       B[2 \pi r]  =  \mu_o  *  \frac{I r^2  }{R^2 }

      B =  \frac{ \mu_o}{2 \pi }  *  \frac{r}{R^2}

When r \ge R

=>     B =  \frac{ \mu_o *  I}{ 2 \pi r }

But when r< R

   B =  [\frac{\mu_o *  I }{ 2 \pi R^2} ]* r

     

4 0
3 years ago
A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy
Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = \sqrt{\frac{3g}{h} }

c )  h = 1.34 m ( 4\pi ^{2} / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= (\pi r^{2} * \frac{h}{3} ) * 2 = \frac{\pi r^{2} h  }{6}

The  part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) =  fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = \frac{\alpha }{2} *\frac{\pi r^{2}  }{h}* a = ( force of gravity ) \alpha * T\frac{r^{2} }{4} * x * g

therefore a = \frac{3g}{h}  * x

angular frequency ( w ) = \sqrt{\frac{3g}{h} }

C ) height of the each cone

\frac{2\pi }{w}  = 1 therefore w = 2\pi

back to the angular frequency : \frac{3g}{h} = 4\pi ^{2}

therefore h = 1.34 m ( 4\pi ^{2} / 3g )

4 0
3 years ago
What acceleration occurs if the friction force in #4 is now 4N?
anastassius [24]

Answer:

you a b1tc h

Explanation:

you're not going anywhere in life. mcdonalds hiring $16 an hour where i live.

4 0
2 years ago
Read 2 more answers
PLEASE HELP URGENT 1. You put 1 gram of salt into 1 liter of water and stir. The resulting liquid is an example of
daser333 [38]
C, is correct.

Have a nice day!
6 0
3 years ago
Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on
Vlad [161]

Answer:

Xcm = 1.95 cm  and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

     R cm = 1/M  ∑ m_{i}  r_{i}

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

     M = m₁ + m₂ + m₃

     M = 140 + 45 + 85

     M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

      R₁ = d / 2 i ^ + d j ^

      R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

      R₂ = 0

Point 3. Right vertex

      R₃ = d i ^

      R₃ = 3.4 i ^ cm

a) The x component of the massage center

      Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

      Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

      Xcm = d / M (m₁ / 2 + m₃)

b)   Let's write the mass center component x

      Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

      Xcm = 238/270

      Xcm = 1.95 cm

c) let's find the component and center of mass

     Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

    Ycm = 1 / M (m₁ d + 0 + 0)

    Ycm = m₁ / M d

d) let's calculate

    Y cm = 1/270 (140 3.4 + 0 + 0)

    Ycm = 1.76 cm

8 0
3 years ago
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