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2 years ago

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A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can

be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?

1 answer:

Pani-rosa [81]2 years ago

7 0

Answer:a) 4+8+24=36

B) 1/4+1/8+1/24=10

C) yu will connect them in parallel connection.

D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.

Explanation:

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At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

Volgvan

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

<em>Lets explain how to solve the problem</em>

- The speed ramp has a length of 121 m and is moving at a speed of

2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

the ground

<em>Lets find your speed on the ground</em>

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

8 0

2 years ago

A(n) ________ has charge but negligible mass, whereas a(n) ________ has mass but no charge.

vfiekz [6]

Electron;Neutron is the correct answer.

6 0

2 years ago

Read 2 more answers

Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi

maria [59]

We really can't tell from the given information.
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.

If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more.

I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.

K.E. = (1/2) (mass) (speed-SQUARED) .

We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.

But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.

That's 4 times as much destructive capability as before.

So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.

5 0

2 years ago

A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

vlada-n [284]

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency $f=7.0\times10^{13}\ Hz$

We need to calculate the reduced mass

Using formula of reduced mass

$\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}$

Where, $m_{H}$ = atomic mass of H

$m_{I}$ = atomic mass of I

Put the value into the formula

$\mu=\dfrac{1\times126.9}{1+126.9}$

$\mu=0.99\ u$

$\mu=0.99\times1.66\times10^{-27}\ Kg$

$\mu=1.643\times10^{-27}\ kg$

We need to calculate the force constant of the spring

Using formula of frequency

$f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}$

$k=f^2\times 4\pi^2\times\mu$

Put the value into the formula

$k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}$

$k=317.8\ N/m$

Hence, The force constant of the spring is 317.8 N/m.

0 0

2 years ago

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.

salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0

1 year ago