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Jet001 [13]
2 years ago
7

A 4 V power source is connected in parallel to four resistors: 1, 2, 2, and 3. Find the total current.

Physics
1 answer:
Keith_Richards [23]2 years ago
4 0

Answer:

Explanation:

Equivalent resistance is 1 / ((1/1) + (1/2) + (1/2) + (1/3)) = 3/7 Ω

I = V/R = 4(7/3) = 28/3 = 9.3 A    

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How planets look like,and maybe How the planets were created
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to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on
Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

8 0
3 years ago
What is x-rays ? and how it's formed ?​
tia_tia [17]

X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.

It has many characteristics such as :

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3 0
2 years ago
Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,
NikAS [45]
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer:  1.3439 x 10⁵ (days²/AU³)

8 0
3 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\\\\

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
3 years ago
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