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Oxana [17]
3 years ago
5

In a classroom demonstration, two long parallel wires are separated by a distance of 2.25 cm. One wire carries a current of 1.30

A while the other carries a current of 3.15 A. The currents are in the same direction. (a) What is the magnitude of the force per unit length (in N/m) that one wire exerts on the other
Physics
1 answer:
Dafna11 [192]3 years ago
4 0

Answer:

Explanation:

Given that,

Current in wire are 1.3A and 3.15A

Distance between wire is d= 2.25cm

d = 2.25/100 = 0.025m

Force per unit length F/l?

Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2).

The field due to I1 at a distance r is given to be

B1 = μo• I1 / 2πr

This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by

F=ILBsinθ

with sinθ=1:

F2=I2 • L •B1

By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2. (Note that F1=−F2.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for B1 into the last equation and rearranging terms gives

F/l = μo• I1• I2 / 2πr

Where μo is constant

μo = 4π×10^7 Tm/A

Then,

F/l = μo• I1• I2 / 2πr

F/l = 4π ×10^-7 × 1.3×3.15/(2π×0.025)

F/l = 3.276×10^-5 N/m

the magnitude of the force per unit length that one wire exerts on the other is 3.276×10^-5 N/m

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Answer:

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A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
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Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

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solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

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w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
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If the distance between two masses is tripled, the gravitational force between changes by a factor of
maw [93]

A. 1/9

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

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6 0
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