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masya89 [10]
4 years ago
13

A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti

vely. What is the value of the effective (rms) voltage of the applied source in the circuit
Physics
1 answer:
Morgarella [4.7K]4 years ago
6 0

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

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A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, a
Natali [406]

Answer:

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

0.0283u₁  + 5.004 x 0   =  1.7(0.0283  +  5.004)

0.0283u₁   =   8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

4 0
3 years ago
A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is
xenn [34]

Answer:

  v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

         Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

        Emf = K + U = ½ m v² - e V₂

Energy is conserved

         Em₀ = Emf

          -eV₁ = ½ m v² - e V₂

           v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0
4 years ago
How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?
lys-0071 [83]

Answer:

  1. Na=1
  2. P=5
  3. Br=7
  4. I=7
  5. Te=6
  6. Sr=2

6 0
2 years ago
n general, which trophic level has the LEAST energy available to it? A) producer B) primary consumer C) secondary consumer D) te
zaharov [31]
D - tertiary consumer

This is because it is the farther up to food chain.  
4 0
3 years ago
The half-life of radium-226 is 1600 years. How many
Alenkinab [10]

Answer:

1600 years half life

3200/1600 is 2 half lifes

first half life takes the sample to 1/2 it's original amount and the second takes it to 1/4 of it's original amount

so 1/4*7gm is 1.75gm

Explanation:

7 0
3 years ago
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