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Keith_Richards [23]

3 years ago

10

Increasing global demands for energy is

1 answer:

siniylev [52]3 years ago

3 0

Bad for the community because we are using up more fossil fuels and energy also creating more smog bad for breathing in every day living.<span />

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As heat is added to water, is it possible for the temperature measured by a thermometer in the water to remain constant?

Mazyrski [523]

Answer:

C. Yes, the water could be changing the phase.

Explanation:

6 0

3 years ago

You are working at a company that manufactures electrical wire. Gold is the most ductile of all metals: it can be stretched into

devlian [24]

Answer: $R=1.424 M\Omega$

Explanation:

Given

mass of gold $m=1.6 gm$

Length of wire $L=2.2 km$

Resistivity of gold $\rho =2.44\times 10^{-8}$

density of gold $=19.3\times 10^3 kg/m^3$

and $mass=volume\times density$

$1.6\times 10^{-3}=volume\times 19.3\times 10^3$

$volume=8.29\times 10^{-8} m^3$

And $Resistance R=\frac{\rho L}{A}$

also be written as

$R=\frac{\rho L^2}{V}$

where $L=length$

$V=volume$

$\rho =resistivity\ of\ gold$

$R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}$

$R=1.42\times 10^{6} \Omega$

$R=1.424 M\Omega$

5 0

3 years ago

OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

Vitek1552 [10]

Answer:

$T=0.02\ s$

$f=50\ Hz$

Explanation:

Given:

minimum amplitude at the start of oscillation cycle, $a_0=20\ unit$

the first maximum amplitude after the start of oscillation cycle, $a_{m1}=100\ units$

Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

<u>So the time taken to complete one cycle of the oscillation:</u>

$T=4\times 0.005$

$T=0.02\ s$ is the time period of the oscillation

<u>We know frequency:</u>

$f=\frac{1}{T}$

$f=\frac{1}{0.02}$

$f=50\ Hz$

6 0

4 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

55. (III) A uniform circular plate of radius 2R has a circular hole of radius R cut out of it. The center of the smaller circle

Aleksandr [31]

Answer:

P = 0.27R from the center

Explanation:

Given,

The radius of the uniform circular plate, R = 2R

The radius of the hole, r = R

The center of the smaller circle from the center is, d = 0.8R

The center of mass of a circular disc with a hole in it given by the formula

P = dr²/R² - r²

Where P is the distance from the center of mass located in the line joining the two centers opposite to the hole.

Substituting the given values in the above equation,

P = 0.8R x R² / 4R² - R²

= 0.27R³/R²

= 0.27R

Hence the center of mass of plate is at a distant P = 0.27R from the center

6 0

3 years ago