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3 years ago

13. Under what condition (if any) does a moving body experience no energy even though there

Physics

1 answer:

djyliett [7]3 years ago

6 0

Answer:

When the Net Forces are equal to 0

Explanation:

Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.

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A uniform charge density of 509 nC/m3 is distributed throughout a spherical volume of radius 6.03 cm. Consider a cubical Gaussia

Artemon [7]

Explanation:

(a) It is known that relation between charge and volume is as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}$

= $509 \times 10^{-9} \times 6.4 \times 10^{-5}$

= $3.26 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 3.68 $N m^{2}/C$

Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 $N m^{2}/C$ .

(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}$

= $509 \times 10^{-9} \times 4.74 \times 10^{-3}$

= $2.41 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 2.72 $N m^{2}/C$

Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 $N m^{2}/C$ .

4 0

4 years ago

,how do charged objects react???quiet urgent

son4ous [18]

Any charged object can exert the force upon other objects ... i think tell me if im right

8 0

3 years ago

Read 2 more answers

As a photon from the Sun's core interacts with an atom or ion, the photon gives up its energy to the particle. But the excited p

Natalka [10]

Answer:

Radiative transport

Explanation:

Radiative transportation takes place extensively in the core of the sun. It produces solar energy which are emitted outward from the core of the sun, and it comprises more than 70% of the sun's heat energy. At the core of the sun, the process of nuclear fusion takes where the hydrogen atoms combines with one another, and forms helium, and during this process, it releases a large amount of energy. The photons continuously interacts with the ions (or atoms) and releases this energy to the particles and the excited particles converts the excess amount of energy into another photon which are carried outward by the process of radiative transport.

8 0

4 years ago

Prove law of conservation of momentum ,??

ziro4ka [17]

Explanation:

$\law \: of \: conservation \: of \: momentum)$