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Answer:

i think its skin..think so this answre help you

The question is incomplete, here is the complete question:

Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]

Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Explanation:

To calculate the energy of one photon, we use the Planck's equation:

[tex]E=\frac{N_Ahc}{\lambda}[/tex]

where,

E = energy of radiation

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\lambda}[/tex] = wavelength of radiation

For 1 photon, the term [tex]N_A[/tex] does not appear

[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex]             (Conversion factor: [tex]1nm=10^{-9}m[/tex] )

Putting values in above equation, we get:

[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]

[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]

Putting values in above equation, we get:

[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]

Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Answer:

Unit of G = m³/kg s²

Explanation:

The force of gravitational attraction is represented by the equation.

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

F is the magnitude of the gravitational attraction on either body

m₁ and m₂ are the masses of the bodies, r is the distance between them. G is the gravitational constant.

We need to find the unit of G.

Rearranging the above formula for G.

[tex]G=\dfrac{Fr^2}{m_1m_2}[/tex]

Putting units of F,m and r such that,

[tex]G=\dfrac{N\times m^2}{kg^2}\\\\\because N=kgm/s^2\\\\G=\dfrac{kg\times m\times m^2}{kg^2\times s^2}\\\\G=m^3/kgs^2[/tex]

Hence, the correct option is (c) m³/kg s².

Answer:

the amplitude of the oscillation is 8 cm

Explanation:

The computation of the amplitude of the oscillation is shown below:

= Length of the spring - length of the equilibrium

= 32 cm - 24 cm

= 8 cm

By deducting the  length of the equilibrium from the length of the spring we can get the amplitude of the oscillation and the same is to be considered

hence, the amplitude of the oscillation is 8 cm

Answer:

r = 31.8

Explanation:

To start with, we make the following assumptions. Let

r be the radius of semicircular end

l be the length of straight section

a be the area inside track, then

p = 2πr + 2l = 400

a = πr² + 2rl

2πr + 2l = 400

l = 200 - πr, substituting for l in area,

a = πr² + 2r(200 - πr)

a = πr² + 400r - 2πr²

a = 400r - πr²

In order to find our minimum area, we differentiate with respect to r and then it equate to zero

da/dr = 400 - 2πr = 0

r = 200/π, substituting for r in equation for l,

l = 200 - πr

l = 0 m

And since this is not possible, l must be 100 m (the minimum), and

r = (200 - 100)/π

r = 100/π = 31.8 m

Answer:

No it is Not because win ratio is differs say Ohio only won 7 out of 7 games and e

was in the playoffs while Alabama won 11 out of 11 games so it is harder to win 11 out of 1a games and it is easier to win 7 out of 7

Explanation:  would like brainliest but u don’t have to

Answer:

B

Explanation:

Just answerd that question!!!

Hope it helped!!!

.

Answer:

D-Enhancing memory

Explanation:

Explanation:

Two atoms with equal masses, m, and equal speed, v, engage in a head on elastic collision. Since, the two atoms both having equal masses and equal but opposite velocity, the total linear momentum of the system is zero. Therefore, conservation of linear momentum to be conserved , the velocities before and after collision are same. Similarly, in elastic collision total kinetic energy is also conserved. The kinetic energy before and after collision are same.

Answer:

Rhythmic Gymnastics

Explanation:

Answer:

[tex]F=1.334*10^-^6[/tex]

Explanation:

From the question we are told that

Mass of Dylan [tex]M_d=75kg[/tex]

Mass of Sarah [tex]M_s=54kg[/tex]

Centers separated by[tex]C=0.45m[/tex]

Generally the equation for solving gravitational force is given by

[tex]F=\frac{G*m_1m_2}{r^2}[/tex]

[tex]F=\frac{6.67*10^-^1^1*75*54}{(0.45)^2}[/tex]

[tex]F=1.334*10^-^6[/tex]

Therefore Gravitational force is given as

[tex]F=1.334*10^-^6[/tex]

Answer:

P = 510 Ns

Explanation:

Momentum is defined as the quantity of motion contained in a body. Mathematically momentum is defined as the product of mass and the velocity of the body. So in order to find out the momentum of the referee after the collision with the football player, we can use the following formula:

[tex]P = mv\\[/tex]

where,

P = Momentum of the referee after the collision = ?

m = mass of the referee = 85 kg

v = speed of the referee after collision = 6 m/s

Therefore,

[tex]P = (85\ kg)(6\ m/s)[/tex]

P = 510 Ns

Answer:

3015 N

Explanation:

From Newton's second law, we know that;

F.t = mv

F = force on the ball= ?

m= mass of the ball

v= velocity

F= mv/t

F= 0.045 × 67/0.0010

F= 3.015/0.0010

F= 3015 N

The force exerted by the racket on the ball is 3015 N.

The force exerted by the racket on the ball is given in the formula below.

⇒ Formula:

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, The force exerted by the racket on the ball is 3015 N

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Answer:

C. 36J

Explanation:

KE = ½ m V²

= ½ × 8 × 3²

= 4 × 9

= 36 J

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

V = 3 + 12

V = 15m/s

Answer:

It looks like a total of 9 meters

Explanation:

the dot on the graph for 1910 sits JUST below the mark for 10 meters

Answer:

look it tilts and light

Answer:

v = 18.5 m /s

Explanation:

The velocity of the jumper will go on increasing until his weight is more than the restoring force created in the bungle.

mg = kx

50 x 9.8 = 100 X

X = 4.9 m

Total fall = 15 + 4.9 = 19.9 m

kinetic energy gained due to fall by 19.9 m

= mgh = 50 x 9.8 x 19.9 = 9751 J

energy lost due to stored elastic energy in bungie

= 1/2 k x²

= .5 x 100 x 4.9²

= 1200.5 J

Remaining kinetic energy = 9751 - 1200.5 = 8550.5 J

1/2 m v² = 8550.5

.5 x 50 x v² = 8550.5

v² = 342

v = 18.5 m /s

Acne is a long-lasting condition that affects the sebaceous glands and hair follicles in the middle layer of the skin.Acne is characterized by clogged sebaceous glands, which produce pimples and cysts.

Which part of the integumentary system is responsible for acne?

        To learn more about acne in adolescents refer

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Answer:

The power transmitted to the pulley is 0.0455 hp.

Explanation:

Given;

mass attached to the rope, m = 5 lb

radius of the pulley, r = 3 in

constant rate of fall of the mass, v = 5 ft/s

acceleration due to gravity, g = 32.2 ft/s²

1 lbf = 32.2 lb.ft/s²

The power transmitted to the pulley is calculated as;

P = Fv

P = (mg)v

[tex]P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2} \ \ = 25 \ \frac{ft.lbf}{s} \\\\[/tex]

in horse power, the power transmitted is calculated as;

[tex]P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s} \ \ = 0.0455 \ hp[/tex]

Therefore, the power transmitted to the pulley is 0.0455 hp.

Answer:

correct answer is d

The pinhole has a size comparable with the laser wavelength, so the pinhole diffracts the passing laser beam.

Explanation:

The researcher does not have a refraction problem since the medium on both sides of the pinhole is the same with the same refractive index, the problem he is having is with the diffraction of the laser beam through the pinhole, let's analyze the diffraction process that is described by the expression

          a sin θ = m λ

where a is the pinlole size,  λ the wavelength of the laser and m an integer.

The laser extends from the maximum of diffraction to the first zero (m = 1) of diffraction

         sin θ =  λ/a

when analyzing this expression we have some interesting cases

* when  λ « a. the sine approaches zero therefore we are in the case of optical geometry, in this case the laser passes through the hole without being diffracted

* when  λ ≈ a.  the sine function has values ​​between 0 and 1, for which a diffraction of the laser beam occurs, which increases the diameter of the same

* when  λ> a.  The laser does not pass through the gap since the sine cannot have values ​​greater than 1

After this analysis, we review the answers to the exercise to find that the correct answer is d

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     [tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

[tex](m/l)=\frac{10}{V^2}[/tex]

Explanation:

From the question we are told that

Speed of a transverse wave given by

[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]

Maximum Tension is [tex]T=10.0N[/tex]

Generally making [tex](m/l)[/tex] subject from the equation mathematically we have

[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]

[tex]v^2=\frac{T}{(m/l)}[/tex]

[tex](m/l)=\frac{T}{V^2}[/tex]

[tex](m/l)=\frac{10}{V^2}[/tex]

Therefore the Linear mass in terms of Velocity is given by

[tex](m/l)=\frac{10}{V^2}[/tex]

Hellow!

For this use the next formula:

Vf = Vo + gt

Initial velocity is zero, so the formula simplificate:

Vf = gt

Data:

Vf = Final velocity = ?

g = Gravity = 9.8 m/s²

t = Time = 12 s

Replacing according our data:

Vf = 9.8 m/s² * 12 s

Vf = 117.8 m/s

The final velocity will be 117.8 meters per second.

Answer:

Time displayed on clock at the front of car is 12:00 noon

Explanation:

According to the question the Time displayed on rear clock of the car displays 12:00 noon.

since car has maximum speed of car is 100 m/s

i.e., they does not move with relativity speed

hence there will be no time dilation in time

Therefore, Time displayed on clock at the front of car is 12:00 noon

Answer:

0.1Ns

Explanation:

Impulse is the product of Force and time

Impulse = Force * Time

Given

Force = 10N

Time = 0.01s

Substitute into the formula

Impulse = 10 * 0.01

Impulse = 10 * 1/100

Impulse = 10/100

Impulse = 0.1Ns

hence the impulse of the hammer is 0.1Ns

Answer:

friction , magnetism etc

Answer:

The answer is below

Explanation:

a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?

Solution:

a) The wavelengths (λ) for standing waves is given by the formula:

[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]

b) The frequency (f) and wavelength (λ) is given by:

fλ = constant

Hence:

[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]

The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:

1. 5.4 meters.

2. 2.7 meters.

3. 1.8 meters.

Given the following data:

To determine the three (3) longest wavelengths for these standing waves:

Mathematically, the wavelength for standing waves is given by the formula:

[tex]\lambda_n = \frac{2L}{n}[/tex]

Where:

Note: n = 1, 2, and 3.

When n = 1:

[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]

When n = 2:

[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]

When n = 3:

[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]

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Answer:

quantity of charge that passes through Anita's window AC, Q = 316800 C

Explanation:

given  data

current I = 11 amps

runs t = 8.0 hours

solution

we get here quantity of charge that passes through Anita's window AC

the quantity of charge passing Q = I × t     .................1

Q =  11 A × ( 8 hr )×  ( 3600 s/hr)

so

quantity of charge that passes through Anita's window AC, Q = 316800 C

The quantity of charge that passes through Anita's window AC, will be

Q = 316800 C

The current is defined as the quantity of charge flowing from the wire per unit time. Also defined as the rate of flow of the charge through any circuit. The formula will be given as:

[tex]\rm I=\dfrac{Q}{t}[/tex]

Now it is given that:

current I = 11 amps

runs t = 8.0 hours

The quantity of charge that passes through Anita's window AC during these 8.0 hours. Will be calculated as:

[tex]Q=I\times t[/tex]

[tex]Q=11\times 8\times 3600[/tex]

[tex]Q=316800 \ C[/tex]

Thus the quantity of charge that passes through Anita's window AC, Q = 316800 C

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Answer:

presence of salt in oceanic water makes 97 of the water on the earth unfit for drinking