A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkinetic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?

Answer:

a) F = 2.2275 * 10^-4 N , upwards

b) F = 2.2275 * 10^-4 N , east to west

c) F = 0

Explanation:

Given:

- The straight wire of length  L = 2.7 m

- The current I in the wire I = 1.50 A

- The magnetic Field B = 0.55 * 10^-4 T

Find:

Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running

a) from west to east

b) vertically upward

c) from north to south

d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Solution:

- If current runs from west to east the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points east. From right hand rule, the direction of force is out of page, so its upward.

- If current runs upward the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points out of page . From right hand rule, the direction of force is out of page, so its east to west.

- If current runs north to south the angle between the magnetic field B is θ = 0°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(0)*10^-4

                           F = 0 N

The magnitude of the force that our planet's magnetic field exerts is mathematically given as

F=1.82* 10^{-4}N

Question Parameter(s):

A straight 2.70 m wire carries a typical household current of 1.50

the earth's magnetic field is 0.550 gauss

Generally, the equation for the  force exerted by the magnetic field   is mathematically given as

F=ILB

Where

B=0.550 G

B= 0.55 *10^{-4}T

In conclusion,force exerted by the magnetic field is

F=(1.50 A)(2.20 m)(0.55* 10^{-4} T)

F=1.82* 10^{-4}N

Read more about Force

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Complete Question

A straight 2.70 m wire carries a typical household current of 1.50 an (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.

Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east

Explanation:

we know that,

linear speed = circumference × revolution per minute

linear speed of belt = 2πr × revolution per minute

now we will compute the linear speed of a belt for 2 inch pulley that is,

linear speed for 2 inch pulley = (2π×2)×( 3 revolutions per minute)  ∵ r =2

                                              = 4π × 3 revolution per minute    (1)

again we will compute the linear speed of a belt for 8 inch pulley,

linear speed of 8 inch pulley = (2π×8) × (x revolution per minute) ∵ r =8

                                            = 16π×x revolutions per minute      (2)

As the linear speed is same for both pulleys. by comparing equations (1) and (2).

                           4π×3 = 16π×x

                              x = 3/4

Thus, the revolutions per minute for the 8 inch pulley is 3/4.

Answer:

(v, 0) or (0, v)

Explanation:

The law of conservation of linear momentum states total initial momentum equal total final momentum.

Total initial momentum = [tex]mv + m\times 0 = mv[/tex]

Total final momentum = [tex]mv_1 + mv_2 = m(v_1+v_2)[/tex]

Equating both,

[tex]mv = m(v_1+v_2)[/tex]

[tex]v = v_1+v_2[/tex]

For an elastic collision, kinetic energy is conserved i.e. total initial kinetic energy = total final kinetic energy

[tex]\frac{1}{2}mv^2 + \frac{1}{2}m\times0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2[/tex]

[tex]v^2 = v_1^2+v_2^2[/tex]

From the first equation, [tex]v_1 = v-v_2[/tex].

Substituting this in the second equation,

[tex]v^2 = (v-v_2)^2+v_2^2[/tex]

[tex]v^2 = v^2-2vv_2 +v_2^2+v_2^2[/tex]

[tex]0 = 2v_2^2-2vv_2[/tex]

[tex]v_2(v_2-v) = 0[/tex]

[tex]v_2 = 0[/tex] or [tex]v_2 = v[/tex]

From [tex]v_1 = v-v_2[/tex],

[tex]v_1 = v-v = 0[/tex]

OR

[tex]v_1 = v-0 = v[/tex]