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Rina8888 [55]
4 years ago
12

How you do this resistor in circuit question?

Physics
1 answer:
lawyer [7]4 years ago
3 0
If Resistors are in series= The equivalent is the sum.

E.g R1 and R2 in series,  R = R1 + R2.

If in Parallel, equivalent is Product/sum.

E.g If R1 and R2 in parallel,  R = (R1*R2)/(R1+R2)

1.)    60 is parallel with 40 and both are then in series with 20.

60//40   = (60*40)/(60+40) = 2400/100 = 24 

Now the 24 is in series with the 20 

R = 24 + 20 = 44 ohms.

2.)    80 is in series with 40 and both are then in parallel with 40.

Solving the series,  R = 80 + 40 =120.

Parallel:   120//40  =     (120*40)/(120+40) =  4800/160 = 30 
Equivalent Resistance = 30 ohms.

3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.

The 1st parallel  =  (100*100)/(100+100) = 10000/200 = 50

The 2nd parallel  =  (50*50)/(50+50)  = 2500/100   = 25.

Solving the series  = 50 + 25 =75 ohms.

Cheers.
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Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
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