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3 years ago

What type of energy system (open, closed or isolated) is represented by the picture and why?

Physics

1 answer:

Vadim26 [7]3 years ago

3 0

<span>An open system can take in and out matter and energy, a closed system takes in and out only energy, and an isolated system does not take in or out energy or matter. In this case it doesn't take out matter, so it should be a closed system.</span>

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Which is the BEST definition of refraction? A) Light or sound waves change direction. B) Light or sound waves bounce off a mediu

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The answer to this is A. this is because, refraction with a light or sound wave changing its direction involve propagation,(in which propagation is the change in direction of a light or sound wave)

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3 years ago

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A scientist needs to determine the average volume of five water samples collected for an experiment. What is

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Explanation:

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3 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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3 years ago

A _________ is the time it takes for the Earth to rotate 360 degrees relative to the background stars.

Tom [10]

The answer is stellar day

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2 years ago

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Calculate the magnitude of electric field strength at a point 3cm from an infinite line of charge of linear density 18 μC/cm sit

valina [46]

Answer:

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

Explanation:

The linear density:

$\begin{gathered} \lambda=18\mu C\text{ /cm} \\ \\ \lambda=\frac{18*10^{-6}C}{0.01m} \\ \\ \lambda=0.0018\text{ C/m} \end{gathered}$

Point r = 3 cm = 3/100 m

r = 0.03 m

The electric field strength is calculated below

$\begin{gathered} E=\frac{\lambda}{2\pi\epsilon_o\epsilon_rr} \\ \\ E=\frac{0.0018}{2\times3.14\times8.85\times10^{-12}\times1.5\times0.03} \\ \\ E=719709237.468\text{ N/C} \\ \\ E=7.2\times10^8\text{ N/C} \\ \end{gathered}$

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

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1 year ago