All categories

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".
______________________________________________________
Explanation:
_____________________________________________________
The formula is: KE = (½) * (m) * (v²) ;
_____________________________________

"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
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".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
___________________________________________________

→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________

→ We divide each side of the equation by: "[(½)* (m)]" ;
___________________________________________________

→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)] ;
______________________________________________________
to get:
______________________________________________________
→ v² = (KE) / [(½)* (m)]

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

→ v = √ { 2*(KE) ] / m } ;

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

and solve for "v".
______________________________________________________

You might be interested in

Pleaseee HElpp!!!!!!!

lutik1710 [3]

Coulomb's Law

Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

Q1 = 3.3x 10^-8 Coulombs

k = 9.0 x 10^9 Newton*m^2/Coulombs^2

Required:

Q2 =?

Formula:

F = k • Q1 • Q2 / d²

Solution:

So, to solve for Q2

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)

Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

Q2 = 3.63636363636 Coulombs

Q2 = 3.64 Coulumbs

6 0

2 years ago

Read 2 more answers

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

How is item A different from Item B?

iris [78.8K]

Explanation:

well there is nothing there and it could be different by diffrent objects, idk

8 0

2 years ago

Read 2 more answers

Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers

bekas [8.4K]

Answer: The correct answer is Option b.

Explanation:

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight

l = length of wire

A = area of cross section

$\Delta l$ = change in length

Hence, the correct answer is Option b.

6 0

3 years ago

Longer wavelengths of light, such as _______, have ________ energy than shorter wavelengths, such as _________

Ilya [14]

Answer:

Micro and radio waves.

Lower energy.

Gamma rays.

Explanation:

The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths.

Ionising radiation os defined as the energy required of photons of a wave to ionize atoms, causing chemical reactions.

The energy of the wave depends on both the amplitude and the frequency. If the energy of each wavelength is a discrete packet of energy, a high-frequency wave will deliver more of these packets per unit time than a low-frequency wave. In summary, the longer the wavelength, the lower the energy to ionise.

The velocity of a wave is directly proportional to the frequency of that wave.

c = f * lambda

Where,

c = velocity of the wave

f = frequency of the wave = 1/time

Lambda = wavelength.

From the above expression, the longer the wavelength, lambda the shorter the frequency.

Examples of waves with longer wavelengths are, micro and radio waves, while radiations with shorter wavelengths like gamma rays.

8 0

2 years ago