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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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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
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Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

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3 years ago
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An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

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       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

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Final. Electron with velocity, just out of the electric field

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          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

5 0
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