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3 years ago

A circuit is built based on this circuit diagram. What is the equivalent resistance of the circuit?

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

7 0

The three resistors are connected to the same points of the circuit, so they are in parallel configuration. The equivalent resistance of 3 resistors in parallel is given by:
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}$
If we plug the values of the resistances into the formula, we find
$\frac{1}{R_{eq}}= \frac{1}{3.0 \Omega}+ \frac{1}{6.0 \Omega}+ \frac{1}{9.0 \Omega}= \frac{6+3+2}{18.0 \Omega} = \frac{11}{18.0 \Omega}$
From which we find the equivalent resistance:
$R_{eq} = \frac{18.0}{11} \Omega =1.6 \Omega$

So, the correct answer is B.

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How?? anyone?!...........

Mariana [72]

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-- find the horizontal and vertical components of F2.

-- find the horizontal and vertical components of F3.

-- add up the 3 horizontal components; their sum is the horizontal component of the resultant.

-- add up the 3 vertical components; their sum is the vertical component of the resultant.

-- the magnitude of the resultant is the square root of (vertical component^2 + horizontal component^2)

-- the direction of the resultant is the angle whose tangent is (vertical component/horizontal component), starting from the positive x-direction.

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2 years 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

Elina [12.6K]

Answer:

0.3 m

Explanation:

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

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(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

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago

What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed

Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above; the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0

3 years ago

A large container is pulled 20 meters with 40 newtons of force. The power exerted is 64 watts. How long did it take to move the

Usimov [2.4K]

A.(20*40)/64=12.5 hope i helped

3 0

3 years ago