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Lemur [1.5K]
3 years ago
13

How will the magnetic field of a straight wire carrying a constant current interact with the needle of a compass? Does the inter

action depend on the relative orientation of the wire and the compass needle? Does the interaction depend on the location of the needle relative to the wire? Does the interaction depend on the distance between the wire and the needle?
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

the needle will direct its North South according to the magnetic field of current carrying wire.

Explanation:

A current carrying wire always has a magnetic field around it, in circular loops. This magnetic field will be either clockwise or anticlockwise depending on the direction of current.

Right hand rule tells the direction. Place the current carrying wire in your right hand with thumb pointing the direction of current. Curl of the fingers tell the direction of current.

When the needle gets in the vicinity of the field, its poles aligns itself with the field. (previous position of the compass needle has no effect on its position in the field). The north pole and south pole will be set in the direction of magnetic field.

The distance between the needle and wire does effect the strength (accuracy) of the needle position. Strong field will create strong deflection of the needle whereas when the distance from wire increases, field weakens, thus the deflection of needle will be weak.

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A small metal bead, labeled A, has a charge of 26 nC . It is touched to metal bead B, initially neutral, so that the two beads s
adell [148]

Answer:

q_A=25.953\ \rm nC

q_B=0.047\ \rm nC

Explanation:

Given:

  • Total initial charge on bead A=26 nC
  • The distance between them=5 cm
  • Magnitude o the force between them =4.4\times10^{-4}

Using coulombs law the force between two charged particle is \dfrac{q_Aq_B}{4\pi \epsilon_0 r^2}

where r is the radial distance between them

According to question we have

4.4\times10^{-4}=\dfrac{q_Aq_B\times9\times10^9}{0.05^2}\\\\4.4\times10^{-4}=\dfrac{q_A(q_A-26\times10^{-9})\times9\times10^9}{0.05^2}\\q_A=25.953\ \rm nC\\q_B=0.047\ \rm nC

Hence the charge on two metal beads is calculated.

5 0
3 years ago
Which method is responsible for the transfer of heat in the atmosphere ?
k0ka [10]

Answer:

B

Explanation: Because conduction is when an object touches another object and convection happens with boiling water or the water cycle, the particles/water fall and rise.

4 0
3 years ago
A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever
kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier (V_s) = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave =  (V_s - V_w)

where:

V_s = velocity of the skier

V_w = velocity of the wave

The wavelength (\lambda) can be written as:

\lambda = (V_s-V_w)T

\lambda = (V_s-V_w) 0.450m ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = (V_s + V_w)

The wavelength (\lambda) can be written as:

\lambda = (V_s+V_w)T

\lambda = (V_s+V_w) 0.342m   ------------------> Equation 2

Equating equation (1) and equation (2) and substituting  V_s  = 14.3 m/s ; we have:

(V_s-V_w) 0.450m  =  (V_s-V_w) 0.342m

0.450m(V_s)-0.450m(V_w)   =  0.342m(V_s)+0.342m(V_w)

Collecting the like terms; we have:

0.450m(V_s) - 0.342m(V_s) =  0.342m(V_w)+0.450m(V_w)

(V_s)(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.108m =  (V_w)0.792m

1.5444m^2/s =  (V_w)0.792m

(V_w) = \frac{1.5444m^2/s}{ 0.792m}

(V_w) = 1.95 m/s

b)

The Wavelength of the wave can be calculated using :  ( \lambda }) = (V_s-V_w) 0.450m

({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)

(\lambda) = (12.35)0.450m

(\lambda)= 5.5575 m

λ ≅ 5.56 m

5 0
3 years ago
g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward t
cupoosta [38]

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v_a = 0.80c

speed of the second spacecraft from earth v_b = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v_a - v_b ) / ( 1 - (v_bv_a / c²) )

we substitute

u' = ( 0.80c - ( -0.60c)  ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) /  ( 1 - ( -0.48c² / c² ) )

u' = 1.4c /  ( 1 - ( -0.48 ) )

u' = 1.4c /  ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c  { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

7 0
3 years ago
PLEASE I NEED THIS QUICKLY
saveliy_v [14]

Answer:

should be A

Explanation:

4 0
3 years ago
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