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madam [21]
3 years ago
12

An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )

Physics
2 answers:
Furkat [3]3 years ago
7 0
KE = 1/2 * m * v2

We have to rearrange this so the subject is mass. (because the question asks for the mass of the object) :

Mass = (2 * KE) / v

Now input the values into this equation to get your answer :

Mass = (2 * KE) / v
Mass = (2 * 480) / 8
Mass = 960 / 8
Mass = 120
bulgar [2K]3 years ago
7 0

Answer:

Mass of an object is 15 kg.

Explanation:

It is given that,

velocity of an object, v = 8 m/s

Kinetic energy of an object, KE = 480 J

Kinetic energy of an object is due to the motion of an object. The mathematical relationship is given by :

KE=\dfrac{1}{2}mv^2

Where m is the mass of an object

m=\dfrac{2\times KE}{v^2}

m=\dfrac{2\times 480\ J}{(8\ m/s)^2}

m = 15 kg

Hence, the mass of an object is 15 kg

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Explanation:

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3 years ago
A 4 kg book and a 7 kg lamp are both in the living room. If the force of gravity between them is 2.99x10^-10 N. How far apart ar
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'G' is the gravitational constant for SI units =  6.67 x10⁻¹¹ newton-meter²/kg²

You said that the force is          <u>2.99 x 10⁻¹⁰</u> = (6.67x10⁻¹¹) (4) (7) / D²

Multiply each side by D² :    2.99 x 10⁻¹⁰ D²  =  (6.67x10⁻¹¹) (28)

Divide each side by  2.99 x 10⁻¹⁰  :          D² = (28) (6.67 x10⁻¹¹) / (2.99 x 10⁻¹⁰)

Take the square root of each side:      D = square root of all that

Stuff it through my calculator:

                               D = square root of  [  (28) (6.67 x10⁻¹¹) / (2.99 x 10⁻¹⁰)  ] =

                                  √ ( 6.2462...) meters²  = 

                                        <em>2.5 meters </em> (rounded)

===================

To get the feel:

-- The book weighs about    8.8  pounds  (big book)
-- The lamp weighs about  15.4  pounds
-- They're about 8.2 feet apart.
-- The strength of the gravitational force
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5 0
2 years ago
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Answer:

A) \,K.E=1.405\times 10^{-20}J

B)\,r_f=0.268\,m

Explanation:

Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s

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B) How close does the proton get to the line of charge?

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Change in voltage is

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