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4 years ago

5

Filtering occurs when _______________. A.)all light is absorbed B.) light reflects off a surface C.) some light is absorbed and

some light is transmitted D.) all light transmitted

1 answer:

stira [4]4 years ago

3 0

Answer:

C.) some light is absorbed and some light is transmitted

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At a constant temperature, the volume of a gas doubles when the pressure is reduced to half of its original value. This is a sta

djyliett [7]

Answer: Boyle’s law

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

As pressure is decreased to half, the volume is increased to doubled.

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles)

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$ (At constant volume and number of moles)

Combined gas Law: combining the three laws:

$PV\propto T$

6 0

4 years ago

Read 2 more answers

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.

Natali5045456 [20]

<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION : lnPso2cl12=-kt+lnPso2cl1 initial partial pressure Pso2cl12 the rate constant k and the time t lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375) so lnPso2cl12=3.002 we take the base e antilog: lnPso2cl12=e3.002 Pso2cl12=20 torr we use the integrated first order rate lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002 we use the same rate constant and initial pressure k=4.5*10-2*s-1 Pso2cl12=375 Pso2cl12=1* so2cl12 Pso2cl12=37.5 torr subtract in Pso2cl12 grom both side lnPso2cl12- lnPso2cl12=-kt ln(x)-ln(y)=ln (x/y) ln (Pso2cl12/Pso2cl20)=-kt we get t -1/k*ln(Pso2cl12/Pso2cl20)=t t=51 s</span>

6 0

4 years ago

An electric current heats a 221 g (0.221 kg) copper wire from 20.0 Â°C to 38.0 Â°C. How much heat was generated by the current?

lorasvet [3.4K]

Answer:

Heat generated by the current = 1547.89 J

Explanation:

We have equation for heat energy H = mCΔT

Mass of copper = 0.221 kg

Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°

ΔT = 38 - 20 = 18°C

Substituting in H = mCΔT

H = 0.221 x 389.112 x 18 = 1547.89 J

Heat generated by the current = 1547.89 J

3 0

4 years ago

In your study unit, a relationship is compared to a car to point out that

mylen [45]

To point out that relationships call for regular maintenance. Cars need regular oil checkups and such and so do relationships.

Hope this helps :)

8 0

4 years ago