A light wave travels through space at a speed of 3 x 108 m/s. If the wavelength of some light wave is 2 x10-3 m, what is the frequency of the wave?


6.67 x 10-12 Hz

1.5 x1011 Hz

6 x105 Hz

2 x 10-3 Hz

A heat engine that uses work to move heat

hope this helps :)

Answer:

D. A heat engine that uses heat to do work

Explanation:

Combustion = Heat

Combustion is used to create motion (work)

Answer:Which statements describe Rutherford’s model of the atom? Select all that apply.

The atom is mostly empty space.

The atom cannot be divided into smaller particles.

Electrons orbit around the center of the atom.

The atom’s positive charge is located in the atom’s nucleus.

The electrons are located within the positive material of the nucleus.

Electron clouds are regions where electrons are likely to be found.

Explanation:

Sample Response: The one major change that occurred was the placement and organization of the electron. Rutherford’s model identified that the electrons were at a distance from the nucleus, Bohr’s model identified that the electrons occurred at levels that related to their available energy, and the modern atomic model shows that electrons are located in a predicted area but cannot be identified in a specific point.

What is the outcome variable in this experiment?

The dependent variable which is the outcome of bubbles produced.

Keesha did an experiment to study the rate of photosynthesis in the water plant Elodea.

My guess would be that since we initially know what the experiment is for, the key phrase is:

Keesha counted the bubbles released from the plant every minute for five minutes (min). She repeated the process two more times.

The outcome of the bubbles that are produced will be the dependent variable.

In conclusion, the key phrase that's vital in knowing what Keesha is looking for was "Keesha counted the bubbles released from the plant every minute for five minutes".

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Answer:

241 kPa

Explanation:

The ideal gas law states that:

[tex]pV=nRT[/tex]

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

[tex]\frac{pV}{T}=nR[/tex]

For a fixed amount of gas, n is constant, so we can write

[tex]\frac{pV}{T}=const.[/tex]

Therefore, for a gas which undergoes a transformation we have

[tex]\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

[tex]p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K[/tex]

So we can solve the formula for [tex]p_2[/tex], the final pressure:

[tex]p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa[/tex]

Answer:

241 kPA

Explanation:

I just did the quest assignment, it was correct

(a) [tex]1.03\cdot 10^{-16} N[/tex], -k direction

First of all, let's find the magnetic field produced by the wire at the location of the electron:

[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]

where

I = 19.0 A is the current in the wire

r = 1.9 cm = 0.019 m is the distance of the electron from the wire

Substituting,

[tex]B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T[/tex]

and the direction is +j direction (tangent to a circle around the wire)

Now we can find the force on the electron by using:

[tex]F=qvBsin \theta[/tex]

where

[tex]q=1.6\cdot 10^{-19}C[/tex] is the electron's charge

[tex]v=3.23\cdot 10^6 m/s[/tex] is the electron speed

[tex]B=2\cdot 10^{-4} T[/tex] is the magnetic field

[tex]\theta[/tex] is the angle between the direction of v and B

In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]. So, the force on the electron is

[tex]F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N[/tex]

The direction is given by the right hand rule:

- Index finger: direction of motion of the electron, +i direction (away from the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)

(b) [tex]1.03\cdot 10^{-16} N[/tex], +i direction

The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:

- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)

(c) 0

In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so [tex]\theta=0, sin \theta=0[/tex], therefore the force on the electron is zero.

Answer:

2f

Explanation:

The relationship between the speed of a sound wave (v), its frequency (f) and its wavelength is

[tex]v=\lambda f[/tex]

For the first wave speed, we have

[tex]v=\lambda f[/tex] (1)

Since the speed of a sound wave depends only on the medium, we can say that the second wave travels at same speed v (because it is still travelling in air, as the first wave). The wavelength of the second wave is [tex]\frac{\lambda}{2}[/tex], so if we call its frequency f', the new equation is

[tex]v=\frac{\lambda}{2}f'[/tex] (2)

And if we equate (1) and (2), we find

[tex]\lambda f = \frac{\lambda}{2}f'\\f'=2f[/tex]

The frequency of the second sound wave is 2f

[tex]\texttt{ }[/tex]

Let's recall the speed of wave formula as follows:

[tex]\large {\boxed {v = \lambda f}}[/tex]

f = frequency of wave ( Hz )

v = speed of wave ( m/s )

λ = wavelength ( m )

Let's tackle the problem!

[tex]\texttt{ }[/tex]

Given:

frequency of first wave = f

wavelength of first wave = λ

wavelegnth of second wave = λ/2

Asked:

frequency of second wave = f' = ?

Solution:

Because the speed of wave depends on the medium, then:

[tex]\texttt{Speed of First Wave = Speed of Second Wave}[/tex]

[tex]v_1 = v_2[/tex]

[tex]\lambda_1 f_1 = \lambda_2 f_2[/tex]

[tex]\lambda f = \frac{1}{2}\lambda f'[/tex]

[tex]f = \frac{1}{2} f'[/tex]

[tex]f' = 2f[/tex]

[tex]\texttt{ }[/tex]

The frequency of the second sound wave is 2f

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Sound Waves

[tex]\texttt{ }[/tex]

Keywords: Sound, Wave , Wavelength , Doppler , Effect , Policeman , Stationary , Frequency , Speed , Beats, Medium, Space

Gravitational Potential energy is the energy stored with regard to position.

the equation for potential energy is as follows

P.E = mass x gravitational acceleration x height from the ground

gravitational acceleration - 9.8 ms⁻²

substituting the values in the equation

P.E = 125 kg x 9.8 ms⁻² x 1500 m

P.E = 1 837 500 J

therefore the potential energy of the diver at 1500 m high is 1 837 kJ

The potential energy of the skydiver from the given height is 1837.5kJ.

Given the data in the question;

Potential energy; [tex]U = \ ?[/tex]

Gravitational potential energy is simply the potential energy an object possesses in relation to another object due to gravity.

It is expressed as:

[tex]U = mgh[/tex]

Where m is the mass of the object, h is its height from ground level and g is acceleration due to gravity ( [tex]g = 9.8m/s^2)[/tex]

Now, to determine the potential energy of the diver, we substitute our values into the expression above.

[tex]U = mgh\\\\U = 125kg\ *\ 9.8m/s^2\ *\ 1500m\\\\U = 1837500kgm^2/s^2\\\\U = 1837500J\\\\U = 1837.5kJ[/tex]

Therefore, the potential energy of the skydiver from the given height is 1837.5kJ.

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(a) 3539 N/m

Hook's law states that:

[tex]F=kx[/tex]

where F is the force applied on the spring, k is the spring constant, x is the stretching of the spring.

In this problem, we have:

[tex]F=mg=(65.0 kg)(9.8 m/s^2)=637 N[/tex] is the force applied (the weight of the fish)

[tex]x=0.180 m[/tex] is the stretching of the spring

Solving the equation for k, we find the spring constant:

[tex]k=\frac{F}{x}=\frac{637 N}{0.180 m}=3539 N/m[/tex]

(b) 0.85 s

The period of oscillation of a spring-mass system is

[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]

In this case,

m = 65.0 kg

k = 3539 N/m

Substituting into the formula,

[tex]T=2 \pi \sqrt{\frac{65.0 kg}{3539 N/m}}=0.85 s[/tex]

(c) 0.37 m/s

The initial elastic potential energy of the spring when the fish is pulled down is:

[tex]U=\frac{1}{2}k\Delta x^2[/tex]

where

[tex]\Delta x = 5.00 cm=0.05 m[/tex] is the stretching of the spring with respect to the initial position

Substituting,

[tex]U=\frac{1}{2}(3539 N/m)(0.05 m)^2=4.4 J[/tex]

The spring reaches its maximum speed when it crosses the equilibrium position, for which [tex]\Delta x=0[/tex], so when all the elastic potential energy has been converted into kinetic energy:

[tex]E=K=\frac{1}{2}mv^2[/tex]

where v is the speed of the fish. Solving for v, we find

[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(4.4 J)}{65.0 kg}}=0.37 m/s[/tex]

The force constant of the spring is  3.53 * 10^3 N/m,  the period of oscillation of the fish is 0.115 s and  the maximum speed it will reach is 0.869 m/s

A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m.

By using the amount of the spring is stretched by the weight of the fish, we can calculate the force  constant [tex]k[/tex] of the spring.

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex] with [tex]v_{max}=\omega A = 2 \pi f A[/tex]

When the fish hangs at rest the upward spring force   [tex]|F_x| = kx[/tex] equals the weight [tex]mg[/tex]of the  fish [tex]f = \frac{1}{T}[/tex].  Therefore the amplitude of the SHM is [tex]0.0500 m.[/tex]

a)Find the force constant of the spring. Express your answer with the appropriate units.

[tex]mg=kx\\k = \frac{mg}{x}  \\k = \frac{65*9.8}{0.180} = 3.53 * 10^3 N/m\\[/tex]

b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units.

[tex]T = 2\pi \sqrt{\frac{m}{k}[/tex]

[tex]T = 2\pi \sqrt{\frac{65 kg}{ 3.53*10^3 N/m} [/tex]

 [tex]T =0.115 s[/tex]

c)What is the maximum speed it will reach?

[tex]v_{max}=\omega A = 2 \pi f A\\v_{max}=\omega A = \frac{ 2 \pi A}{T} \\v_{max}=\omega A = \frac{ 2 \pi 0.05 m}{0.115}\\v_{max}=\omega A = 0.869 m/s[/tex]

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When action potentials reach the end of the axon, they stimulate opening of Ca2+ channels, causing a release of neurotransmitters to the post-synaptic cell. How does and impulse propagate down the axon? The stimulus causes a start of the action potential and it moves down the axon without the ions moving down

Answer:

= 3.869 × 10^-6 J/m³

Explanation:

Intensity is given as W/m^2 which is equivalent to J/ (s*m^2)

Speed of light is 2.99792 × 10^8 m/s.

Therefore;

Electromagnetic energy per cubic meter = Intensity/speed of light

      = 1160 J/ (s*m³)/ 2.99792 × 10^8 m/s.

      = 3.869 × 10^-6 J/m³

Answer:

1) F=ma  

a=(mh*g)/(7/5ms+1/2md+mh)

2) same answer as 1 because it is a system

3)a=alpha*R

alpha=a/Rd (use a from 1)

4) similar to 3

alpha=a/Rs

5)F=ma

T=ms*a

T=7/5ms*a (a from 1)  

6)T=mh(g-a)

7)d=1/2at^2 (position formula - starting position and velocity are zero)

t=sqrt(2d/a) (use a from 1)

8)vf^2=vi^2 +2ad (initial velocity = 0)

vf=sqrt(2ad)

9)omega=v/Rs

Explanation:

The question involves a complex system of objects linked by a string and requires an understanding of principles like tension, angular and linear acceleration, and gravity. Using Newton's laws of motion and rotation, as well as a free-body diagram, one can identify linear and angular accelerations and string tension. The time taken for the hoop to fall can be calculated using an equation of motion.

This is a complex system of objects attached to a string and it involves a number of different physical principles such as gravity, tension, angular acceleration, linear acceleration, and moment of inertia.

The linear acceleration of the hoop and the sphere, as well as the angular acceleration of the disk pulley and the sphere, can be identified using the Second Law of Motion (Net Force = mass x acceleration) and Newton's second law for rotation (Net Torque = Moment of Inertia x Angular Acceleration).

The tension in the string can be obtained using free-body diagrams for each object and writing equations for gravitational force and tension.

The time taken by the hoop to fall a certain distance can be calculated using the equation of motion. However, this is a complicated problem and might be better solved using numerical methods or advanced physics concepts.

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b is the correct answer

Answer:

c- interact with charged particles.

Explanation:

When a charged particle is placed in the field of the other charged particle then it will experience electromagnetic force on it.

As we know that force on a charged particle due to some other electric field is given by

F = qE

here we know

q = charge

E = electric field intensity

Similarly when a charged particle is placed in external magnetic field then the force on that moving charge is given by

[tex]F = q(v\times B)[/tex]

so here if charge is moving in the magnetic field due to some other system then the force on it is given by above equation.

So here we can say that electromagnetic force is present when electromagnetic fields

c- interact with charged particles.

Answer:

612.2 m

Explanation:

The work needed to lift the object is equal to its increase in gravitational potential energy:

[tex]W=\Delta U=mg \Delta h[/tex] (1)

where

m = 1.0 kg is the mass

g = 9.8 m/s^2

[tex]\Delta h[/tex] is the vertical distance

The power provided is

[tex]P=0.10 kW = 100 W[/tex]

In one minute (t = 1 min = 60 s), the work provided is

[tex]W=Pt=(100 W)(60 s)=6000 J[/tex]

Substituting this into (1) and solving for [tex]\Delta h[/tex], we find

[tex]\Delta h=\frac{W}{mg}=\frac{6000 J}{(1.0 kg)(9.8 m/s^2)}=612.2 m[/tex]

The amount of energy needed to power a 0.10-kW bulb for one minute is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. The distance is calculated using the formula Work = Force * Distance. The distance is found to be 0.61 meters.

The amount of energy needed to power a 0.10-kW bulb for one minute is 6 Joules. This energy is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. To calculate the distance, we can use the formula:

Work = Force * Distance

In this case, the force is equal to the weight of the object, which is 1.0 kg * 9.8 m/s² (acceleration due to gravity). So, the distance is equal to:

Distance = Work / (Force * Acceleration due to gravity)

Substituting the values, we get:

Distance = 6 J / (1.0 kg * 9.8 m/s²) = 0.61 meters

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Answer:

The current halves

Explanation:

The relationship between voltage, current and resistance in a circuit is given by Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage

R is the resistance

I is the current

We can rewrite the formula as

[tex]I=\frac{V}{R}[/tex]

we see that I is directly proportional to V and inversely proportional to R.  In this problem, V is held constant while R is doubled:

[tex]R'=2R[/tex]

so, the new current in the circuit will be

[tex]I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}I[/tex]

So, the current halves.

Answer:

2940 kg m/s

Explanation:

First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:

[tex]v_y = v_0+gt[/tex]

where

v0 =0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration due to gravity

t = 3 s is the time of the fall

Substituting into the formula, we find

[tex]v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s[/tex]

And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:

[tex]p=mv=(100 kg)(29.4 m/s)=2940 kg m/s[/tex]

The motorcycle's momentum when hitting the ground is calculated using the formulas for momentum and velocity. With a mass of 100 kg and a velocity calculated as 29.4 m/s, the motorcycle's momentum is 2940 kg.m/s.

In this case, we are asked to find the momentum of the motorcycle when it hits the ground. The momentum (p) of an object is calculated by the formula p=mv where m is the mass and v is the velocity. The velocity of the motorcycle can be calculated using the formula v=gt, where g is the acceleration due to gravity (9.8 m/s2) and t is the time. So, v=9.8*3 = 29.4 m/s. Therefore, the momentum of the motorcycle is p=100*29.4 = 2940 kg.m/s. Compared to the typical automobile with a momentum of 1400*15 = 21000 kg.m/s, the motorcycle has significantly less momentum due to its smaller mass and velocity.

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Answer: Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer.

Explanation:

Answer: d

Explanation:

Answer:

4. At twice the distance the field strength will be 1/4 of the original value

Explanation:

The magnitude of the electric field produced by an isolated poitn charge is

[tex]E=k\frac{q}{r^2}[/tex]

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, the distance from the charge is doubled from 4 m to 8 m:

[tex]r'=2 r[/tex]

Therefore, the new electric field will be

[tex]E'=k\frac{q}{(r')^2}=k\frac{q}{(2r)^2}=k\frac{q}{4r^2}=\frac{1}{4}E[/tex]

So, the field strength is 1/4 of the original value.

By doubling the distance between two charges the field strength will be    [tex]\dfrac{1}{4} th[/tex]  of the original value.

The formula for the magnitude of the electric field is given by  

[tex]E=k\dfrac{q}{r^2}[/tex]

Here

k is the Coulomb's constant

q is the charge

r is the distance from the charge

It is given in the problem that the distance from the charge is doubled from 4 m to 8 m:

[tex]r'=2r[/tex]

So to find the new electric field

[tex]E'=k\dfrac{q}{r^2} = k\dfrac{q}{(2r)^2} =\dfrac{1}{4} k\dfrac{q}{r^2} =\dfrac{1}{4} E[/tex]

Thus By doubling the distance between two charges the field strength will be    [tex]\dfrac{1}{4} th[/tex]  of the original value.

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When light travels from air into water, it slows down and bends towards the normal, due to the higher refractive index of water compared to air. This change in speed and direction is known as refraction. It's explained by Snell's law of refraction.

When light travels from air into water, it notices a change in speed, due to the difference in the refractive indices of the two mediums. This change in speed causes the light to change direction, a phenomenon we know as refraction. Since the index of refraction for water (approx 1.33) is higher than that of air (approx 1.0003), the light slows down and bends towards the normal upon entering the water from the air. This is based on Snell's law of refraction, expressed as n₁ sin 0₁ = n₂ sin 0₂, where n represents the index of refraction and 0 represents the angle of incidence. Thus, the higher index of refraction in water than in air causes the beam of light to refract towards the normal.

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A pure substance because it cannot be separated into parts

A process occurs in which a system's potential energy increases while the environment does work on the system. The kinetic energy of a system decreases while its potential energy and thermal energy are unchanged.

Kinetic energy is the energy which is present in the body of an object which is under motion. Kinetic energy of an object transforms into potential energy and in the form of work when the object changes its state of motion to rest.

When a process occurs, the potential energy of the system increases while the environment does some work on the system. In this case, the kinetic energy of the object decreases while the potential energy and thermal energy of the system remains unchanged. This is because kinetic energy is used in doing work.

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Your question is incomplete, most probably the complete question is:

A process occurs in which a system's potential energy increases while the environment does work on the system. Does the system's KINETIC energy increase, decrease, or stay the same?

sound waves even if no one is around to hear it.

A falling tree in a forest creates a sound wave. This is caused by the energy of the tree disturbing the atoms in the air, causing them to vibrate and create a wave. Sound can only be produced where there is matter like air or water.

When a tree falls in a forest, it creates a disturbance in the air around it. This disturbance of air particles initiates a sound wave. Sound waves are a type of wave that our ears have evolved to detect.

As the tree makes impact, the energy transfers to the atoms in the air, causing them to vibrate. These vibrations then spread from particle to particle, creating a wave that travels through the air. This is the definition of sound in physics, regardless of whether there is anyone present to hear it or not.

However, it's important to note that these sound waves would only be produced in an environment that has matter (like air or water) for the wave to travel through. In space, where there is a vacuum, no sound would be produced if a tree fell because there would be no matter to disturb.

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Answer:

E) its speed and wavelength decrease, but its frequency stays the same

Explanation:

First of all, the frequency of a light wave does not depend on the medium, while wavelength and speed do. Therefore, the frequency remains costant.

In particular, the speed of light in a medium is given by:

[tex]v=\frac{c}{n}[/tex]

where c is the speed of light in a vacuum and n is the index of refraction. From the formula, we see that v and n are inversely proportional: so, when the light moves into a material with higher index of refraction, its speed decreases.

Moreover, speed is related to wavelength by

[tex]v=\lambda f[/tex]

where [tex]\lambda[/tex] is the wavelength and f is the frequency. Since the two quantities are directly proportional, this means that since the speed decreases, the wavelength decreases as well.

So, the correct choice is

E) its speed and wavelength decrease, but its frequency stays the same

When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same.

[tex]\texttt{ }[/tex]

Let's recall Snell's Law about Refraction as follows:

[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]

where:

n₁ = incident index

θ₁ = incident angle

n₂ = refracted index

θ₂ = refracted angle

[tex]\texttt{ }[/tex]

In this problem, we will use this following formula:

[tex]\boxed{ \frac{n_1}{n_2} = \frac{v_2}{v_1} }[/tex]

where:

n₁ = incident index

v₁ = incident speed

n₂ = refracted index

v₂ = refracted speed

[tex]\texttt{ }[/tex]

From above formula we could conclude that the speed of light is inversely proportional to the index of refraction. Therefore, when light goes from one material into another material having a HIGHER index of refraction , its speed will decrease.

[tex]\texttt{ }[/tex]

As we know that :

[tex]\boxed{v = \lambda f}[/tex]

Speed of light is directly proportional to the wavelength of light.

If the speed of light decreases , then wavelength of light will also decrease.

[tex]\texttt{ }[/tex]

When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Light

C. How much sleep each student gets

It says directly that you are trying to test how the amount of sleep a student gets affects their performance. If that is the case, it would make sense to change the hours or amount of sleep each student gets before taking the test.

Answer:

3.73 m/s^2

Explanation:

The period of a simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)

where

L is the length of the pendulum

g is the gravitational acceleration on the planet

The pendulum in the problem has length

L = 0.640 m

and makes 10 oscillations in 26.0 s; it means that its frequency is

[tex]f=\frac{10}{26 s}=0.385 Hz[/tex]

And so its period is

[tex]T=\frac{1}{f}=\frac{1}{0.385 Hz}=2.6 s[/tex]

So now we can solve equation (1) using L=0.640 m and T=2.6 s, so we can find the value of g on the planet:

[tex]g=(\frac{2\pi}{T})^2L=(\frac{2\pi}{2.6 s})^2 (0.640 m)=3.73 m/s^2[/tex]

Answer:

-0.805 m

Explanation:

The x-component of a vector is given by:

[tex]v_x = v cos \theta[/tex]

where

v is the magnitude of the vector

[tex]\theta[/tex] is the angle of the vector with respect to the positive x-direction

In this problem we have

v = 0.888 m

[tex]\theta=205^{\circ}[/tex]

so we have

[tex]v_x = (0.888 m)(cos 205^{\circ})=-0.805 m[/tex]

This is chemical energy the substance that makes it glow are chemicals.

Answer:

Chemical .

Explanation:

On Halloween, you take a glow stick, crack the capsule inside and shake it until it glows. This is an example of light energy being created from ___________________ energy

From the principle of the conservation of energy which states that energy can neither be created nor destroyed but can converted from one form to another

Phenyl oxalate ester is responsible for the luminescence in a glow stick.When it reacts with hydrogen peroxide, the liquid inside a glow stick to glow.

therefore we can say that chemical energy is been converted to light energy

the glow lights are by no means radioactive. it can be made to glow again by putting it in a freezer.

1) D. The collisions between molecules are inelastic

Explanation:

The kinetic theory of the gases describe the property of the gases by looking at microscopic level. At these level, some assumptions are made on the motion/collisions of the molecules of the gas:

- Molecules move by random motion

- The volume of the molecules is negligible compared with the volume of the gas

- The molecules obey Newton's laws of motion

- The intermolecular forces between the molecules are negligible except during the collisions

- Collisions between molecules are elastic

Therefore, the following statement

D. The collisions between molecules are inelastic

is wrong.

2) [tex]\sqrt{2}[/tex]

The kinetic energy Ek of a gas is directly proportional to its absolute temperature T:

[tex]E_k = \frac{3}{2}kT[/tex]

where k is the Boltzmann's constant. However, the kinetic energy depends on the square of the average velocity of the particles, [tex]v^2[/tex]:

[tex]E_k = \frac{1}{2}mv^2=\frac{3}{2}kT[/tex]

where m is the mass of the particles. This means that the velocity is proportional to the square root of the temperature:

[tex]v \propto \sqrt{T}[/tex]

So, if the temperature of the gas is doubled, the average speed increases by a factor [tex]\sqrt{2}[/tex], and the ratio v2/v1 is

[tex]\frac{v_2}{v_1}=\sqrt{2}[/tex]

Well , its a easy problem  

By ohms law,  

we have V= IR  

Where V is the voltage applied across the wire and R is the resistance and I is the current  

=> R = V/I  

=> R = 12/0.3 = 40 ohm

The  resistance of the copper wire will be 400 ohms. Resistance is found as the ratio of the voltage and electric current.

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The given data in the problem is;

V is the voltage= 12 V

I is the current = 0.30A

R is the resistance

The resistance of the circuit from the Ohm's law is found as;

[tex]\rm V= IR \\\\ R = \frac{V}{I} \\\\\ R = \frac{12}{0.30} \\\\\ R =400 \ ohm[/tex]

Hence, the resistance of the copper wire will be 400 ohms.

To learn more about the resistance, refer to the link;

https://brainly.com/question/20708652

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Given the data in the question;

Electricity used; [tex]E = 1000kWh = 36*10^8J[/tex]

Time; [tex]t = 1 month = 2.592 * 10^6s[/tex]

Voltage;  [tex]V = 120V[/tex]

Calculate the average rms current to the house.

First we determine the power

Power is the amount of energy transferred per unit time:

[tex]Power = \frac{E}{t}[/tex]

So we substitute in our values

[tex]Power = \frac{36*10^8J}{2.592*10^6s}\\\\Power = 1.388 * 10^3 J/s \\\\Power = 1.388 * 10^3 W[/tex]

Next we Calculate the Current

From Ohms Law:

[tex]P = I * V[/tex]

We substitute in our values

[tex]1.388*10^3W = I \ *\ 120V \\\\I = \frac{1.388*10^3W}{120V} \\\\I = 11.57 W/V\\\\I = 12A[/tex]

Therefore, the average rms current in the power line to the house is 12A

The resistance of a household

From Ohm's Law:

[tex]R = \frac{V}{I}[/tex]

We substitute in our values

[tex]R = \frac{120V}{12A}\\\\R = 10ohms[/tex]

Therefore, The average resistance of a household is 10Ω.

Learn more; https://brainly.com/question/13037925