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3 years ago

Phân biệt các đặc điểm khác nhau giữa chất rắn, chất lỏng

Physics

1 answer:

vitfil [10]3 years ago

6 0

Answer:

şen çal kapimi turkish drama

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Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.

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I think the answer is “greenhouse effect”

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3 years ago

How are the electric field lines around a positive charge affected when a second positive charge is near it?

Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

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3 years ago

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an object has a displacement of + 12 m in 6 seconds if the average velocity is found using the equation change in position / ela

rosijanka [135]

Answer:

+2m/s

Explanation:

average velocity = displacement traveled / total time taken

= +12m/ 6s

= +2 m/s

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3 years ago

A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to

Alborosie

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

$\epsilon=Blv$

The electric field in the conductor is given by :

$E=\dfrac{\epsilon}{l}\\\\E=\dfrac{Blv}{l}\\\\E=Bv$

It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

6 0

3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago