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3 years ago

0.125 C of charge flow out of a

Physics

1 answer:

zimovet [89]3 years ago

3 0

Answer:

Resistance = 252.53 Ohms

Explanation:

Given the following data;

Charge = 0.125 C

Voltage = 5 V

Time = 6.3 seconds

To find the resistance;

First of all, we would determine the current flowing through the battery;

Quantity of charge, Q = current * time

0.125 = current * 6.3

Current = 0.125/6.3

Current = 0.0198 A

Next, we find the resistance;

Resistance = voltage/current

Resistance = 5/0.0198

Resistance = 252.53 Ohms

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Answer:

$1.1486813808\times 10^{-7}\ T$

34.46 V/m

Explanation:

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I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

$B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T$

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The electric field intensity is 34.46 V/m

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3 years ago

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

Maslowich

Answer:

.a = 849.05 m / s²

Explanation

The centripetal acceleration is

a = v² / r

Linear and angular velocity are related

v = w r

Angular velocity and frequency are related by

w = 2π f

Let's replace

a = w² r

a = 4π² f² r

Let's reduce to the SI system

f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

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a = 4π² 14.45² 0.103

.a = 849.05 m / s²

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3 years ago

A hydrogen atom has a diameter of about 12.7 nm, express this diameter in centimeters.

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1 meter = 1e9 nm
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2 years ago

Read 2 more answers

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.

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

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