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12345 [234]
3 years ago
6

Two resistors of 2 ohms each are connected in parallel, another resistor of 1 ohm is connected in series with the parallel combi

nation, and the resistive combination is connected to a 2 volt source. How much power is dissipated in either one of the parallel resistors?
Physics
1 answer:
insens350 [35]3 years ago
3 0

Answer:

The power dissipated in either one of the parallel resistors is 2 V

Explanation:

Given;

two parallel resistors, R₁ and R₂ = 2 ohms

The total resistance of the Two resistors of 2 ohms connected in parallel is;

R_T = \frac{R_1R_2}{R_1+R_2} = \frac{2*2}{2+2} = \frac{4}{4} = 1 \ ohm

when connected to another resistor  of 1 ohm in series, the total resistance becomes;

Rt = R₁ + R₂

Rt = 1 + 1 = 2 ohms

Current in the circuit, I = voltage / total resistance

                                      = 2 /2 = 1 A

the overall circuit has been resolved to series connection, and current flow in series circuit is constant.

Power = I²R

Thus,  power dissipated in either one of the parallel 2 ohms resistors is;

Power = I²R = (1)² x 2 = 2 V

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A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

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3 years ago
Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
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<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

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3 years ago
At speeds over 30 mph, you should maintain a following distance of at least ________ behind the vehicle ahead of you.
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At speeds over 30 mph, you should maintain a following distance of at least <u>three full seconds</u> behind the vehicle ahead of you.

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To count the full three seconds you can use the technique of counting the Mississippis as follows: Mississippi one, Mississippi two, Mississippi three.

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An accident is an unexpected event that generally causes damage, injury or negative consequences.

Learn more about accident at: brainly.com/question/28070413

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1 year ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
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Answer:

0.3 m

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Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

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v = 2 m/s

k = 30000 N/m

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(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
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