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3 years ago

Which statement best describes why the universe could be considered an isolated system?

1 answer:

7 0

<span> In the Universe , all matter and energy contained; no energy or matter </span>
<span>leaves the Universe. </span>

You might be interested in

Pressure and volume changes at a constant temperature can be calculated using

V125BC [204]

Answer:

Option A, Boyle's law

Explanation:

The complete question is

Pressure and volume changes at a constant temperature can be calculated using

a. Boyle's law. c. Kelvin's law.

b. Charles's law. d. Dalton's law.

Solution

In Boyle’s law, the gas is assumed to be ideal gas and at constant temperature. With these two conditions fixed, Boyle’s established that volume of gas varies inversely with the absolute pressure.

The basic mathematical representation of this phenomenon is as follows -

$P \alpha \frac{1}{V}$

$P = \frac{k}{V} \\PV = k$

Where P is the pressure of ideal gas, V is the volume and k is the constant of proportionality.

Hence, option A is correct

3 0

3 years ago

Diego hypothesizes that the Earth wobbles on its axis in a pattern that is repeated every 20,000 years.

DedPeter [7]

Answer:

D. make a model

Explanation:

3 0

3 years ago

Human muscles have an efficiency of about

ryzh [129]

Answer: the answer is B. 18 to 26%

Explanation: I took the practice test on Edge and I got it right :)

8 0

3 years ago

Numerical Problems

Evgesh-ka [11]

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

$20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ]$

4 0

3 years ago

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

4 0

4 years ago