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Nostrana [21]
3 years ago
9

Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors?

What about a "balanced bridge?

Physics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

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Kinematics
leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

v_{f} = v_{i}+(a*t)

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2} =  v_{i} ^{2}+(2*a*x)

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

v_{f} = v_{i}-a*t

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2}  = v_{i} ^{2} -2*a*x3\\

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0
2 years ago
Solving elastic collisions problem the hard way
vova2212 [387]

Answer:

<h2>Solving elastic collisions problem the hard way</h2><h3 />

Explanation:

perfect drawing

4 0
3 years ago
What are the functions of the female reproductive system?
gulaghasi [49]

The female reproductive system is designed to carry out several functions. It produces the female egg cells necessary for reproduction, called the ova or oocytes. The system is designed to transport the ova to the site of fertilization.

7 0
2 years ago
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given that,

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We need to calculate the value of v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, put the all values into the acceleration 's formula

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is a= 0.3296\ \hat{k}\ m/s^2

7 0
3 years ago
Read 2 more answers
Which of the following statements are true for magnetic force acting on a current-carrying wire in a uniform magnetic field? Che
qaws [65]

Answer:

The following statements are correct.

1. The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

2. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

3. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

Wrong statements:

1. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic field lines.

Explanation:

6 0
2 years ago
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