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kodGreya [7K]
3 years ago
13

Consider a loop of wire whose plane is horizontal and that carries a current in the clockwise direction when viewed from above.

If we were to represent the current loop as a bar magnet or magnetic dipole, in what direction would the north pole be pointing?
Physics
1 answer:
Mars2501 [29]3 years ago
7 0

Answer:

Downwards into the plane

Explanation:

Solution:-

- This is a conceptual application of  hand rule. We will place our palm fingers open vertical to a plane surface. Then curl our fingers in and naturally point the thumb.

- The direction of curl of fingers denotes the direction of of current flow in the coil. Which in our case is "clockwise direction". We will orient/invert our right hand palm in such a way that we curl our fingers in clockwise fashion. Then stick the thumb out to give us the direction of magnetic field or North pole end. In our case the the thumb points downwards into the plane denoting that the magnetic field within the loop is also acting downwards into the plane.

- The bar magnet would be placed in such a way that North pole is pointing downward into the plane in the direction of magnetic field and end up at south pole pointing up out of the plane.

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A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca
Sonbull [250]

Answer:

The correct solution is:

(a) 1.375\times 10^{-2} \ J

(b) 4.43\times 10^{-3} \ C

(c) 1.42\times 10^{-3} \ F

(d) 178.57 \ \Omega

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

P_{avg}=55 \ mW

       =55\times 10^{-3} \ W

Average voltage,

V_{avg}=3.1 \ V

Now,

(a)

⇒ E=P_{avg}\times \zeta

On substituting the values, we get

⇒     =55\times 10^{-3}\times 0.25

⇒     =1.375\times 10^{-2} \ J

(b)

⇒ E=Q\times V_{avg}

then,

⇒ Q=\frac{E}{V_{avg}}

On substituting the values, we get

⇒     =\frac{1.375\times 10^{-2}}{3.1}

⇒     =4.43\times 10^{-3} \ C

(c)

⇒ C=\frac{Q}{V}

⇒     =\frac{4.43\times 10^{-3}}{3.1}

⇒     =1.42\times 10^{-3} \ F

(d)

As we know,

⇒ R=\frac{1}{4C}

⇒     =\frac{1}{4\times 1.42\times 10^{-3}}

⇒     =\frac{1000}{5.6}

⇒     =178.57 \ \Omega

5 0
2 years ago
A 4 kg bowling ball rolls at a speed of 5 m/s on the roof of a building that is 30 meters tall. What is its kinetic energy?
klasskru [66]

ans: B

Kinetic energy = \frac{1}{2} mv^{2}

= \frac{1}{2} (4)(5^{2} )

= 50J

ps. the height will only affect its potential energy, not kinetic.

3 0
3 years ago
How can you determine the type of radiation an atom emits as its nucleus decays
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Gamma raidiation and beta and alpha particles
7 0
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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
marshall27 [118]

Answer:

The resultant velocity is  v_t=10 knots

Explanation:

Apply the law of conservation of momentum

     M_L *v_L + M_f * V_f = (M_L + M_f) v_t

Where M_L is the mass of the Luxury Liner = 40,000 ton

            v_L is the velocity of Luxury Liner = 20 knots due west

            M_f mass of freighter = 60,000

           v_f is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

         v_f = 0 \ knots

So the equation would be

              M_L *v_L = (M_L + M_f) v_t

Substituting values

            40000*20 = (40000+ 60000)v_t_w

Where v_t_w the final velocity due west

Making v_t_w the subject

          v_t_w = \frac{40,000* 20}{(40000 + 60000)}

                = 8 \ knots

Apply the law of conservation of momentum toward the the north direction          

          v_L = 0 \ knots

So the equation would be

           M_f *v_f = (M_L + M_f) v_t_n

Where v_t_n the final velocity due north

     Making v_t_n the subject

          v_t_n = \frac{60,000* 10}{(40000 + 60000)}

                = 6 \ knots

The resultant velocity is

       v_t = \sqrt{v_t_w^2 + v_t_n^2}

            = \sqrt{8^2 +6^2}

           v_t=10 knots

8 0
3 years ago
Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal
elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

  • W = F║d
  • where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

  • W = (87 N)(4.5 m)
  • W = 391.5 J  

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0
2 years ago
Read 2 more answers
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