Okay! Last question was a warm-up question. And now to get your brain thinking some more, how about another one?:

Your high-fidelity amplifier has one output for a speaker of resistance 8 Ω. How can you arrange two 8-Ω speakers, one 4-Ω speak

ANTONII [103]

Answer:

(a) 8Ω (b) Ratio = Parra/P8 ohm = 1

Explanation:

Solution

Recall that,

An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω

Now,

(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker

The Upper arm is : 8 ohm, 8 ohm

The Lower arm is : 12 ohm, 4 ohm

The Requirement is = (16 x 16)/(16 + 16) = 8 ohm

(b) compare your arrangement power output of with the power output of a single 8-Ω speaker

The Ratio = Parra/P8 ohm = 1

8 0

3 years ago

a certain car travels 20 km east then turns south for 13 km finally the car turns east again for 6 km

aliya0001 [1]

a^2+b^2=c^2

20km^2+13km^2=c^2

400km^2+169km^2=c^2

23.853720...km

7 0

2 years ago

Read 2 more answers

A trained eye in the dark for an extended period of time may pick up a light stimulus from a light source, at the lowest radiate

frez [133]

Answer:

#_photons = 30 photons / s

Explanation:

Let's start by finding the energy of a photon of light, let's use the Planck relation

E = h f

the speed of light is related to wavelength and frequency

c = λ f

we substitute

E = h c /λ

E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹

E₀ = 3.978 10⁻¹⁹ J

now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second

#_photons = 1 photon (E / Eo)

#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹

#_photons = 3.0 10¹

#_photons = 30 photons / s

3 0

2 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

2 years ago

Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c

dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

$\omega = 2\pi\frac{24}{20}$

$\omega = 2.4\pi rad/s$

Now we know that the tangential speed of the chain is given as

$v = r \omega$

so we have

$v = (3 cm)(2.4\pi)$

$v = 7.2 \pi cm/s$

$v = 432\pi cm/min$

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0

3 years ago