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3 years ago

11

Voltage has been described as similar to the water in the water pipes in your home. What other analogy can you think of that is

similar to voltage?

1 answer:

Kisachek [45]3 years ago

3 0

When describing voltage, current, and resistance, a common analogy is a water tank. In this analogy, charge is represented by the water amount, voltage is represented by the water pressure, and current is represented by the water flow.

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Ice that formed thousands of years ago is often found to contain tiny bubbles of gas. this gas came from __________. ice that fo

motikmotik

Atmosphere

Atmospheric gas from prehistoric eras is found trapped in glaciers in the form of bubbles. These gas bubbles are the basis of studying ice cores as they provide us with accurate estimates of the conditions of past climates. The bubbles allow us to determine the composition of atmospheric air, such as the carbon dioxide and methane concentrations, as well as allow us to determine air temperatures in the past.

8 0

3 years ago

Arisa [49]

Answer:

A. nuclear fusion reactions

C. it's still hot from the big bang

Explanation:

The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:

Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
Some heat accreted during the early formation of the earth and has not been lost till this day.
Heating due to friction

These are some of the sources of the earth's internal heat.

3 0

2 years ago

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

5 0

2 years ago

On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters

dimulka [17.4K]

Answer:

$d_1 = 16 d$

Explanation:

As we know that initial speed of the fall of the stone is ZERO

$v_i = 0$

also the acceleration due to gravity on Mars is g

so we have

$d = v_i t + \frac{1}{2}gt^2$

now we have

$d = 0 + \frac{1}{2}g t^2$

now if the same is dropped for 4t seconds of time

then again we will use above equation

$d_1 = 0 + \frac{1}{2}g(4t)^2$

$d_1 = 16(\frac{1}{2}gt^2)$

$d_1 = 16 d$

7 0

3 years ago

Read 2 more answers

WHOEVER GET IT RIGHT GETS 50 POINTS

AVprozaik [17]

Answer:

ANSWER BELOW I

I

V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

3 0

3 years ago