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3 years ago

How do you find the speed of an object given its mass and kinetic energy (what is the formula)?

1 answer:

8 0

v = √ { 2*(KE) ] / m } ;

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;

and solve for "v".
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Explanation:
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The formula is: KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
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we arrange the formula ;
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(KE) = (½) * (m) * (v²) ;  ↔ (½)*(m)* (v²) = (KE) ;
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→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ;
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→ [ (½)*(m)*(v²) ] /  [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
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to get:
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→ v² =   (KE) / [(½)* (m)]

→ v² = 2 KE / m
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Take the "square root" of each side of the equation ;
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  → √ (v²) = √ { 2*(KE) ] / m }
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→ v = √ { 2*(KE) ] / m } ;

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];

and solve for "v".
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svetoff [14.1K]

1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..

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2 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

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Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

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2 years ago

A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h

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Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

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3 years ago

A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves.

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To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.

Mathematically its definition can be given as:

$d sin\theta =m\lambda$

Where

d = Width of the slit

$\theta =$ Angle between the beam and the source

m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)

Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:

$\lambda = d$

$\lambda = 20cm = 20*10^{-2}m$

Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:

$f = \frac{v}{\lambda}$

Here the velocity is equal to the speed of light and the wavelength to the value previously found.

$f = \frac{3*10^8}{0.2}$

$f = 1.5Ghz$

Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz

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3 years ago

A. At what point in it's motion is the kinetic energy of the end of a pendulum greatest? b. At what point is the potential energ

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The kinetic energy will be greatest at the bottom of the swing motion.

The potential energy will be greatest at the highest position of the swing.

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In an oscillating pendulum, the potential energy and gravitational kinetic energy are constantly changing. The potential and kinetic energies are maximal at extreme and intermediate positions, respectively.

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