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3 years ago

explain why if charge cannot be created or destroyed, electrically neutral objects can become electrically charged

1 answer:

5 0

Charge is actually either excessive number or electrons or shortage of them.
as you can't destroy the electron or create it - you can't destroy or create charge

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How can you increase the intensity of sound waves

VladimirAG [237]

Answer

The intensity of a sound wave depends on the pressure of the wave,density of the medium and speed of sound in the medium. Higher density and higher sound speed both give a lower intensity. and may be it is because that sound wave is more characterize by wavelength than frequency..explanation

Explanation:

As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times greater. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.

5 0

3 years ago

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$