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3 years ago

Which example involves the transformation of chemical energy directly into light energy?

2 answers:

6 0

B) a glow stick , glow sticks are made up of certain chemicals and when they fuse together they start to glow which would be considered as light energy

AleksandrR [38]3 years ago

6 0

The answer is B) A glow stick.

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A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a

xxTIMURxx [149]

Answer: $y_{com}=0.707 m$

$v_{com}=3.713 m/s$

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by first stone in 430 ms

$y_1=ut+\frac{at^2}{2}$

$y_1=0+\frac{g(0.43)^2}{2}$

$y_1=0.9069 m$

Distance traveled by stone 2 in t=430-59=371 ms

$y_2=ut+\frac{at^2}{2}$

$y_2=0+\frac{g(0.0.371)^2}{2}$

$y_2=0.674 m$

velocity of first stone after t=0.43 s

$v_1=u+at$

$v_1=0+9.8\times 0.43=4.214 m/s$

velocity of second stone after t=0.371 s

$v_2=u+at$

$v_2=0+9.8\times 0.371=3.63 m/s$

Position of Center of mass of system

$y=\frac{y_1m_1+y_2m_2}{m_1+m_2}$

$y=\frac{0.9069\times m+0.674\times 6m}{m+6m}$

$y=\frac{4.95m}{7m}=0.707 m$

Velocity of COM

$v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}$

$v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}$

$v_{com}=3.713 m/s$

6 0

4 years ago

When light shines through atomic hydrogen gas, it is seen that the gas absorbs light readily at a wavelength of 91.63 nm. What i

Artist 52 [7]

Answer:

D) 21

Explanation:

When gas absorbs light , electron at lower level jumps to higher level .

and the difference of energy of orbital is equal to energy of radiation absorbed.

Here energy absorbed is equivalent to wavelength of 91.63 nm

In terms of its energy in eV , its energy content is eual to

1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .

Electron is lying in lowest or first level ie n = 1.

Energy of first level

= - 13.6 / 1² = - 13.6 eV.

Energy of n th level = - 13.6 / n². Let in this level electron has been excited

Difference of energy

= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)

13.6 / n² = 13.6 - 13.57 = .03

n² = 13.6 / .03 = 453

n = 21 ( approx )

4 0

3 years ago

Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field

swat32

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

7 0

3 years ago

Read 2 more answers

A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d

Usimov [2.4K]

<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn prevents the car from slipping down the curve.

So,

If the car doesn't hit the gas then the car will fall down from the curve as the Centripetal force will exceed the Centrifugal force of the car.

However, if the car doesn't hit the brake then the car will maintain it's position on the flat curve track as the centrifugal force will counter the effect of centripetal force directed towards the center.

3 0

3 years ago

Read 2 more answers

20. A 0.145 kg mass of tungsten at 130.0 °C is placed in a 0.502 kg of water at 22.0 °C. The mixture reaches equilibrium at 28.6

zepelin [54]

MaCa∆Tb = MbC∆Tb
Where m is the mass of a compound
 C is the specific heat of a compound
∆T is the change of temperature of the compound
 (0.502 kg water)(4180 j/kg C)( 28.6 – 22 C) = ( 0.145 kg tungsten) ( C ) ( 130 – 28.6 C)
Solve for C
C = 941.9 J /kg C

8 0

3 years ago