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3 years ago

7

the resistor of values 6 ohm,6 ohm are connected in series and 12 ohm are connected in parallel. the equivalent resistance of th

e circuit is

1 answer:

andrezito [222]3 years ago

6 0

Answer:

The equivalent or total resistance of the circuit is 6

Explanation:

6 &6 are in series

6+6=r

r= 12

1/Rtotal= 1/12+1/2

1/Rt=2/12=1/6

Rt=6

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Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

<span>Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span> $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

<span>Like this:
</span>
$I_{T} = \frac{E}{ R_{T} }$

$I_{T} = \frac{20}{ 260 }$
$I_{T} = 0,076923076...$

$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
<span>The intensity of the total electric current
</span> $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

4 0

3 years ago

A crate falls from an airplane flying horizontally at an altitude of 2,000 m.

Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

7 0

3 years ago

A football player who weighs 550 N stands indoors wearing her football boots. The boot’s

EleoNora [17]

Answer:

183333 Pa

Explanation:

The weight of the football player is : 550 N ,thus the force the player exerts on the floor is 550 N

The area of blades in contact with the floor is = 30cm² = 0.003 m²

Pressure = Force / Area

Pressure = 550 / 0.003

Pressure = 183333 Pa

3 0

3 years ago

A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

Novosadov [1.4K]

Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C

8 0

3 years ago

Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

3 years ago