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Lesechka [4]
3 years ago
7

A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field o

f 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Physics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Explanation:

- To find the direction of the conventional current in the wire you use the following formula:

\vec{F}=i\vec{l}\ X\ \vec{B}       (1)

i: current in the wire = ?

F: magnitude of the magnetic force on the wire = 15.1N

B: magnitude of the magnetic field = 6.1T

l: length of the wire that is affected by the magnetic field = 0.45m

The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).

The direction of the current must be in the +y direction (+^j). In fact, you have:

^j X ^k = ^i

The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):

F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

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A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the
Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

4 0
3 years ago
A defibrillator passes 14.0 A of current through the torso of a person for 0.0300 s. How much charge moves in coulombs?
krek1111 [17]
<h2>Answer:</h2>

4.2 C

<h2>Explanation:</h2>

The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.

i.e

Q = I x t

Where;

I = current = 14.0A

t = time taken  = 0.0300s

Substituting the values of I and t into the equation above gives

Q = 14.0 x 0.0300

Q = 4.2 C

Therefore quantity of charge moving is 4.2 C

4 0
3 years ago
If an object is not accelerating, it can exist in what 2 other states of motion?
dezoksy [38]

Answer:

Friction is a force that acts in a direction opposite to the motion of the moving object.

Friction will cause a moving object to slow down and finally stop.

7 0
3 years ago
A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0
3 years ago
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How is air pressure related to elevation?
exis [7]
Air pressure decreases with altitude

hope that helps you!
6 0
3 years ago
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