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asambeis [7]
3 years ago
13

Be sure to answer all parts. Without stratospheric ozone (O3), harmful solar radiation would cause gene alterations. Ozone forms

when the bond in O2 breaks and each O atom reacts with another O2 molecule. It is destroyed by reaction with Cl atoms formed when the C―Cl bond in synthetic chemicals breaks. Find the wavelengths of light that can break the C―Cl bond and the bond in O2. Give your answers in scientific notation. C―Cl × 10 m O2 × 10 m
Physics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

for C-Cl  bond \lambda_1=3.66\times 10^{-9}\ m

for O-O bond \lambda_2=2.43\times 10^{-9}\ m

Explanation:

  • We have the bond enegy for C-Cl, BE_1=327\ kJ.mol^{-1}= 327000\times 1.66\times 10^{-22}\ J\ per\ molecule
  • bond enegy for C-Cl, BE_2=494\ kJ.mol^{-1}= 494000\times 1.66\times 10^{-22}\ J\ per\ molecule

<u>Now as we know the energy of electromagnetic waves is given by:</u>

E=h.\nu

here E = BE

<u>Then for C-Cl:</u>

327000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_1

\nu_1=8.18733\times 10^{16}\ Hz

Now wavelength:

\lambda_1=\frac{c}{\nu_1}

\lambda_1=\frac{3\times 10^8}{8.18733\times 10^{16}}

\lambda_1=3.66\times 10^{-9}\ m

<u>For O2:</u>

494000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_2

\nu_2=1.236863\times 10^{17}\ Hz

Now wavelength:

\lambda_2=\frac{c}{\nu_2}

\lambda_2=\frac{3\times 10^8}{1.236863\times 10^{17}}

\lambda_2=2.43\times 10^{-9}\ m

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A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai
patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

<u>t = 4.79 s </u>

The equation for velocity is as follows:

v = v0  + a* t

Where:

v = velocity

v0 =  initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>

b) The time interval was calculated above, using the equation of  the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

<u>t = 1.55 s</u> ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = <u>6.34 s</u>

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

<u>The car lands 24 m from the base of the cliff.</u>

PHEW!, it was a very complete problem :)

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3 years ago
A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

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2 years ago
If element x has 44 protons how many electrons does it have?
Jlenok [28]
The correct answer is '44'.

In fact, the atom of an element has an equal number of protons and electrons: therefore, if element X has 44 protons, it should have 44 electrons as well.

For curiosity: the chemical element with 44 protons and 33 electrons is Ruthenium.
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Coherent light with wavelength 540 nm passes through narrow slits with a separation of 0.370 mm .Part AAt a distance from the sl
77julia77 [94]

Answer:

The phase difference between waves 31.7 rad

Explanation:

Given :

Wavelength \lambda = 540 \times 10^{-9} m

Separation between two slit d = 0.370 \times 10^{-3} m

Angle \theta = 25°

From the formula of phase difference,

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Where \delta = phase difference

\delta = \frac{2 \times \pi \times 0.370 \times 10^{-3} }{540 \times 10^{-9} } \sin 25

\delta = 31.7 rad

Therefore, the phase difference between waves 31.7 rad

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