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Aleonysh [2.5K]

2 years ago

12

The energy transfer diagram shows energy transfer in an MP3 player. Useful energy is transferred away from the MP3 player by lig

ht and what else

1 answer:

OLEGan [10]2 years ago

4 0

Answer:

heat and sound.

Explanation:

Though some would argue that the heat is not useful. I guess it depends on if your hands are cold.

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The number that is used to show the value of one currency compared to another is called the __________. A. trade rate B. currenc

Brrunno [24]

Answer:

c is the answer

Explanation:

8 0

2 years ago

A carnot engine operates between two heat reservoirs at temperatures th and tc. an inventor proposes to increase the efficiency

Sonja [21]

The efficiency of the first Carnot engine is
n1 = 1 - Th/T
The efficiency of the second Carnot engine is
n2 = 1 - T/Tc
The total efficiency of the engines put in series is
n = 1 - Th/Tc
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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

An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has

kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is $1.50\times 10^{-3}\text{ C}$ .

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

$I(t) = I_0e^{-t/\tau}$

Here, $I_0$ is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

$I = \dfrac{V}{R_\text{eff}}$

<em>V</em> is the emf of the battery and $R_\text{eff}$ is the effective resistance.

In this question, $R_\text{eff}$ = 10.0 Ω + 25.0 Ω = 35.0 Ω

$I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}$

(b) The maximum charge is given

<em>Q</em> = <em>CV</em>

where <em>C</em> is the capacitance of the capacitor

$Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}$

3 0

2 years ago

Read 2 more answers

An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{4}{9.81}$

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0

2 years ago