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natta225 [31]
3 years ago
5

engineers who design battery operated devices suck as sell phones and MP3 players try to make them as efficient as possible. An

engineer test a cell phone and finds that the batteries supply 10,000 J of energy to make 5,500 J of output energy in the form of sound and light for the screen. How efficient is the phone
Physics
1 answer:
ICE Princess25 [194]3 years ago
8 0
If the 5,500 J of sound and light is the ONLY useful output
from the phone, then the phone's efficiency is

                   (5,500J / 10,000J)  =  0.55  =  55% .

But the test engineer forgot one little minor almost insignificant detail.
As a test engineer myself, I'd say that he needs to turn in his laptop
and soldering iron, and think about changing his career to a job for
which he may be better suited, like 8 hours a day in a highway toll-booth.  

What about that little radio transmitter and receiver inside the phone,
that maintain digital RF communication with the nearest cell tower ?
Without that microscopic radio transceiver ... plus 30 or 40 apps
that are always running unless you shut them off ... the device in your
pocket is essentially a flat rock with one side that sometimes glows.
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Mark all the mesons a) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino
Sloan [31]

Answer:

None

Explanation:

Subatomic particles are the particles which are very smaller than the atoms. Elementary particles are the examples of subatomic particles.

Elementary particles are the particles without any sub-structure which means they are not composed of other particles.

The elementary particles are classified into three categories which are discussed below:

(1) Quarks: up, down, top, bottom, strange, and charm.

(2) Leptons: muon, muon neutrino, electrons, electron neutrino,  tau, tau neutrino.

(3) Bosons:  Z bosons, W bosons, Higgs, Gluon, photons.

Mesons are the particles which compose one quark and one anti quarks.

Therefore, in the given list there is no meson.

3 0
3 years ago
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
3 years ago
A motorcycle accelerates from a stop at a rate of 4m/s 2 for 20 seconds. It then continues at a
8_murik_8 [283]

After 20 s, the motorcycle attains a speed of

\left(4\dfrac{\rm m}{\mathrm s^2}\right)(20\,\mathrm s)=80\dfrac{\rm m}{\rm s}

and it continues at this speed for the next 40 s. So at 45 s, its speed is 80 m/s.

4 0
3 years ago
Which of the following is equivalent to Planks constant
Alinara [238K]

Answer:

6.62607004 x 10^(-34)m²kg/s

Explanation:

This is the constant that shows the value of energy of a photon in relation to it's frequency.

Please let me know if you want this explained further!

Thanks!

8 0
3 years ago
Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b
sergeinik [125]

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

<u>Explanation:</u>

Work, W = 2 J

Initial distance, x_{1} = 30 cm

Final distance,  = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8  cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

7 0
3 years ago
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