What is the direction of velocity in uniform circlar motion

Physics

1 answer:

Alborosie3 years ago

5 0

Answer:

$the \: direction \: of \: velocity \: in \: uniform \: circular \: \\ motion \: is \: alway \: \boxed{tangential \: to \: the \: circle}. \\ that \:i s \: \underline{ \boxed{the\: direction \: of \: this \: velocity \:i s \: always \: changing}}.$

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g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

AnnyKZ [126]

Answer:

Time, $t=1.07\times 10^{-23}\ s$

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel $3.2\times 10^{-15}\ m$ before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

$t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s$

So, the time interval is $1.07\times 10^{-23}\ s$

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3 years ago

which of the glass lenses above, when placed in air, will cause rays of light (parallel to the central axis) to converge?

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A virtual image is one that cannot be produced on a screen and is formed when the image is upright in relation to the object. When an item is positioned within one focal length of a converging lens, a virtual image is created. It creates an enlarged image of the object on the same side of the lens as the image. It serves as a magnifier.

Learn more about the convex lens here:

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1 year ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

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To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

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Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

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