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uysha [10]
3 years ago
6

Two balls have the same mass. One ball moving along the positive x axis and the other ball moving along the positive y axis unde

rgo perfectly inelastic collision. What can you say about the motion of the balls after the collision?
A. Stick and move together
B. Separate immediately
C. First ball stops while other starts moving
D. The two balls travel off in opposite directions
Physics
2 answers:
Paladinen [302]3 years ago
7 0
In a completely inelastic collision, the two objects stick together after the collision.
Semenov [28]3 years ago
3 0
<h3><u>Answer;</u></h3>

A. Stick and move together

<h3><u>Explanation;</u></h3>
  • <u><em>A perfect inelastic collision is a type of collision in which in which part of kinetic energy is converted to other form of energy during the collision.</em></u>
  • In an elastic collision on the other hand, is a type of collision in which both conservation of momentum and conservation of kinetic energy are observed.
  • <em><u>During perfectly inelastic collision, the objects or bodies involved stick together and move as a single object after the collision.</u></em> These type of collisions result to in loss of maximum kinetic energy, which is converted into heat or in the work done in deforming the objects involved.

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It would be B I think
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Why can the suns rays cause a burn but the light from a flashlight cannot harm your skin?​
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3 years ago
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Which characteristics accurately describe the inner planets? Check all that apply. small large made of gas made of rock few moon
evablogger [386]

Answer:

Small, made of rocks, made of few moons.

Explanation:

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3 years ago
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Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius R = 7.5 cm. Assume that the powder
Ipatiy [6.2K]

Answer:

A) E =\frac{\rho r}{2\epsilon}

B) v = 58.7923\times 10^4 V

Explanation:

a) using Guass law

\oint E.dA = \frac{q_{enclosed}}{\epsilon_o}

EA = \frac{q_{enclosed}}{\epsilon_o}

E =  \frac{q_{enclosed}}{A \epsilon_o}

HereA is = 2\pi rL

E =  \frac{q_{enclosed}}{(2\pi rL) \epsilon_o}

Volume charge density is given as

\rho = \frac{q_{enclosed}}{volume}

net charge is given as

q_{enclosed} = \rho \times volume

therefore E =  \frac{ \rho \times (L\pi r^2)}{(2\pi rL) \epsilon_o}

VOLUME  =  L\pi r^2

After solving electric field equation we get

E =\frac{\rho r}{2\epsilon}

b) electric potential difference is given as

v_{wall} - v = - \int_{r}^{R} Edr

0 - v = - \int_{r}^{R} E dr

v = \int_{r}^{R} Edr

= \int_{r}^{R} (\frac{\rho r}{2\epsilon}) dr

= \frac{\rho}{4\epsilon} (R^2 - r^2)

at r = 0

v = \frac{- 3.7 \times 10^{-3} \times 0.075^2}{4\times (8.85\times 10^{-12}}

v = 58.7923\times 10^4 V

7 0
3 years ago
The constant of proportionality between charge and voltage is the:________.
Brut [27]

Answer:

The constant of proportionality between charge and voltage is the capacitance.

Explanation:

The capacitance (C), measured in farads, is the ratio of charge to voltage, that is:

C = \frac{Q}{V}

Where:

Q - Charge, measured in coulombs.

V - Voltage, measured in volts.

If charge is clear, it is easy to concluded that capacitante is a constant of direct proportionality.

Hence, the correct answer is B.

4 0
3 years ago
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