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Elodia [21]
3 years ago
15

Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

từ trường tại tâm O của hình vuông.
Physics
1 answer:
disa [49]3 years ago
4 0

Answer:

The net magnetic field ta the center of square is1.36\times10^{-4} T.

Explanation:

Current, I = 12 A , side ,a = 10 cm =  0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B  \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 +  sin 45  \right )\\\\B = 3.4\times 10^{-5} T

Net magnetic field at the center of the square is

B' = 4 B

B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T

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Action-reaction forces are not balanced forces because of what?
lara [203]

Oh but they are !

Newton's 3rd law of motion says that for every action, the <em><u>re</u></em>action is
equal and opposite.  That's as balanced as you can get.


4 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
What requirement must a force acting on a object satisfy in order for the object to undergo simple harmonic motion?
viktelen [127]

Answer:

Simple harmonic motion is the movement of a body or an object to and from an equilibrium position. In a simple harmonic motion, the maximum displacement (also called the amplitude) on one side of the equilibrium position is equal to the maximum displacement.

The force acting on an object must satisfy Hooke's law for the object to undergo simple harmonic motion. The law states that the force must be directed always towards the equilibrium position and also directly proportional to the distance from this position.

6 0
3 years ago
REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part
sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

    W_{fr} = ΔEm

    W_{fr} = Em_{f} -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

    Em_{f} =W_{fr} + Em₀

 

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

    ½ m v² = -fr d + mgh

    v = √(-fr d 2/m + 2 gh)

    v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0
3 years ago
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a
finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$  after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y $

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

          = 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0
2 years ago
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