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Alina [70]
3 years ago
13

Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius,

from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor
Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

Let the radius is r. And outer is R.

Use the formula of current density

I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A

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The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bump
lukranit [14]

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

8 0
3 years ago
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

3 0
3 years ago
A large raindrop-the type that lands with a definite splat-has a mass of 0.0014 g and hits your roof at a speed of 8.1 m/s. a. W
DENIUS [597]

Answer

given,

mass of the drop, m = 0.0014 g

speed of the drop, u = 8.1 m/s

a) Change in momentum is equal to impulse

   final velocity of the drop, v = 0 m/s

   J = m ( v - u )

   J = 0.0014 x 10⁻³ x ( 0 - 8.1 )

   J = -1.134 x 10⁻⁵ kg.m/s

impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s

b) time, t = 0.37 m s

   impact of force = ?

  we know

     J = F x t

   1.134 x 10⁻⁵ = F x 0.37 x 10⁻³

     F =  0.031 N

the magnitude of the force of the impact is equal to F =  0.031 N

5 0
3 years ago
An electron and a proton are fixed at a separation distance of 973 nm. Find the magnitude and the direction of the electric fiel
Vilka [71]

Answer:

The magnitude is: |E|=6084.1\: N/C

The direction of E is in the negative x-direction.

Explanation:

The electric field equation is:

E=k\frac{Q}{r^{2}}

Where:

  • Q is the charge (we can choose the electron or the proton)
  • r is the distance (in our case is at the midpoint 973/2 nm)
  • k is the Coulomb constant (9*10^{9}\: Nm^{2}C^{-2})

Using the electron charge (e = -1.6*10^{-19}\: C)

E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}

The magnitude is:

|E|=6084.1\: N/C

The direction of E is in the negative x-direction.

I hope it helps you!

6 0
3 years ago
Which does a reference point provide? Select 2 correct choices.
erastova [34]

Answer:

1 and 4

Explanation:

6 0
3 years ago
Read 2 more answers
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