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liubo4ka [24]
3 years ago
15

Công thức tính quãng đường đi được của chuyển động thẳng biến đổi đều là:

Physics
1 answer:
Archy [21]3 years ago
6 0

At time t = 0 the velocity is v1. Therefore C1 = v1 and C2 = x1. Equations (1), (2), (3), and (4) fully describe the motion of particles, or bodies experiencing rectilinear (straight-line) motion, where acceleration a is constant.

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Name three bright Saturn ring features, and explain why they are so bright.
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The F Ring, the Cassini Division, and the C Ring are bright ring features. They are bright due to the low concentration of materials within them, which allows sunlight to shine through.


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What is the Coriolis Effect?
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An effect whereby a mass moving in a rotating system experienced a force acting perpendicular to the direction of motion and to the axis of rotation.

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a business that sells products to teens would most likely create a website with a title ending in what suffix?
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3 0
3 years ago
Read 2 more answers
A conductor differs from an insulator in that a conductor.
Zina [86]

Answer:

has free electrons

Explanation:

A conductor has free electrons while an insulator does not. Free electrons are electrons which are not bounded tightly to their parent atoms, and are free to move given the right conditions (ie. a strong EM field).

4 0
3 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

5 0
3 years ago
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