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3 years ago

Help please, i am lost.

Physics

1 answer:

AURORKA [14]3 years ago

4 0

Answer:

nas wr hdrtuendnetyje because

Explanation:

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Help wit these questions someone.

romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, $\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...$

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}$

R₂ + 2 = 12

R₂ = 10Ω

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}$

Req = 1.7Ω

7 0

3 years ago

A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is conne

lora16 [44]

Answer:

Energy stored in the capacitor will be $1.835\times 10^{-6}J$

Explanation:

We have given edge length of capacitor = $1.2\times 10^2cm=1.2m$

So area A = 1.2×1.2 = 1.44 $m^2$

Separation is given as d 1 mm = $10^{-3}m$

We know that capacitance is given by $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F$

Voltage is given as V = 12 volt

We know that energy stored is given by $E=\frac{1}{2}CV^2=\frac{1}{2}\times 12.744\times 10^{-9}\times 12^2=1.835\times 10^{-6}J$

3 0

4 years ago

Read 2 more answers

Before designing the filter, one must understand the relationship between the output voltage of the circuit and the frequency. F

Kamila [148]

Answer:

Option b. is correct

Explanation:

An RLC electrical circuit consists of constituent components: a resistor (R), an inductor (L), and a capacitor (C). A resistor, an inductor, and a capacitor are connected in series or parallel.

The impedances of the circuit elements depend on the frequency.

Both impedance magnitudes decrease when the frequency increases

5 0

3 years ago

A charge of 24 C pass a certain point in a conductor in 10 s. What is the size of current in the conductor?

allsm [11]

Answer:

write the name of any five districts of nepal

6 0

3 years ago

Read 2 more answers

1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin

Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0

3 years ago