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vitfil [10]
3 years ago
8

Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

rent of 4.36 A .Is the force per meter exerted on the 4.36-A wire greater than, less than, or the same as the force per meter exerted on the 2.79-A wire
Physics
1 answer:
Dafna1 [17]3 years ago
3 0

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

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The answer of this question is B. 22x + 20
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Instruments on board the trmm (tropical rainfall measuring mission) satellite show 3d images of very tall rain columns called __
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How do you test hydrogen gas ​
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4 0
2 years ago
A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of
kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0
3 years ago
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