Ask question

Ask question

All categories

3 years ago

9

You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will ha

ve a value that is

1 answer:

belka [17]3 years ago

7 0

Answer:

<h2>The equivalent is 6R/11</h2>

Explanation:

We know that the equivalent resistance of resistors connected in parallel is expressed as

$\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\$

the L.C.M is 6R

$\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}$

You might be interested in

The angle of incidence and the angle of reflection are measured from the_____, the perpendicular to the surface.

kogti [31]

I believe it is angles

6 0

3 years ago

Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent

Harman [31]

Answer:

earth

Explanation:

The formula for the orbital period of the moon is given by

$T = 2\pi \sqrt{\frac{r}{g}}$

As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.

As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.

5 0

3 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

5 0

3 years ago

A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the

Nezavi [6.7K]

Answer:

$F = 371.738\,N$

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

$(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )$

$F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}$

The average force acting on the ball during the collision is:

$F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}$

$F = 371.738\,N$

3 0

3 years ago

Read 2 more answers

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press

Ghella [55]

Answer:

$\mathbf{F_1=4.41*10^4\ N}$

$\mathbf{F_2 = 1.176*10^5 \ N}$

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

$\sum F = 0$ and $\sum \tau = 0$

From the image, expressing the forces through the y-axis, we have:

$F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}$

Also, let the force $F_1$ be the pivot and computing the torque to determine $F_2$ :

Then:

$F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P$

$F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}$

$F_2 = 117600 \ N$

$\mathbf{F_2 = 1.176*10^5 \ N}$

For the force equation:

$F_1+F_2=1.617*10^5 \ N;$

where:

$F_2 = 1.176*10^5 \ N$

Then:

$F_1+1.176*10^5 \ N=1.617*10^5 \ N$

$F_1=1.617*10^5 \ N-1.176*10^5 \ N$

$F_1=44100\ N$

$\mathbf{F_1=4.41*10^4\ N}$

8 0

3 years ago