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yanalaym [24]
3 years ago
10

5. An infinite wire is 3.00 cm away from and parallel to a 10 cm long wire, and each has a current of 2.50 A in the same directi

on (for example the top wire of the rectangle in Fig. 4). (a) What is the magnetic field from the infinite wire at the position of the other wire
Physics
1 answer:
Step2247 [10]3 years ago
3 0

Answer:

B=1.04*10^{-4}T

Explanation:

An infinite wire produces a magnetic field that can be computed with the formula:

B=\frac{\mu_0 I}{2\pi r}

where m0 is the magnetic permeability of vacuum (4pi*10^{-7}Tm/A), I is the current of the wire (2.50A) and r is the perpendicular distance to the wire in which B is calculated (3cm=3*10^{-2}m). By replacing we obtain:

B=\frac{(4\pi*10^{-7}Tm/A)(2.50A)}{3*10^{-2}m}=1.04*10^{-4}T

hope this helps!!

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Small paragraph explaining how how potential and kinetic energy are related
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Answer:

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment.

Explanation:

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If one 9V battery is used in a circuit with a total resistance of 39Ω, what is the current in the circuit?
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Answer:

Using V= IR

I= 0.2307 Ampere

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what percentage of North American adults may be functioning below their potential due to prolonged exposure to stress
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Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
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