Why does refracting light bend when it enters a denser or less dense medium?

One mystery that was exhibited by the motions of the sun and a few other sky objects was the apparent retrograde motion of the planets.

In ancient times, people observed that the planets (including Mars, Jupiter, and Saturn) would sometimes appear to move backward in the sky for a few weeks or months before resuming their forward motion.

This retrograde motion was difficult to explain using the geocentric model of the universe, in which the Earth was considered to be the center of the cosmos and all the other celestial bodies revolved around it. It wasn't until the heliocentric model was proposed, with the sun at the center of the solar system, that this mystery could be explained as a result of differences in the orbital speeds and distances of the planets.

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The client is most likely to receive the information about weather condition of the atmosphere like any active heavy storm from the weather satellite.

Weather satellite is a satellite that gives us the information about the atmospheric conditions of the planet by revolving around it.

The weather satellites are used in order to be aware of the upcoming heavy Storm, heavy rains, floods and Tsunami etc.

When a client is using a weather satellites it is most likely that he will first receive the information about an upcoming heavy storm that may hit the city.

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Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

a=0.1388m/s²

Answer:

The acceleration of dog is 0.15 m/s².

Step-by-step explanation:

Given :

To Find :

Using Formula :

\(\star{\underline{\boxed{\sf{\purple{{v}^{2} = {u}^{2} + 2as}}}}}\)

Solution :

Substituting all the given values in the formula to find acceleration :

\({\longrightarrow{\sf{{v}^{2} = {u}^{2} + 2as}}}\)

\({\longrightarrow{\sf{{6}^{2} = {3}^{2} + 2 \times a \times 90}}}\)

\({\longrightarrow{\sf{{6 \times 6}= {3 \times 3} + 2 a \times 90}}}\)

\({\longrightarrow{\sf{36=9 + 180a}}}\)

\({\longrightarrow{\sf{36 - 9 = 180a}}}\)

\({\longrightarrow{\sf{27 = 180a}}}\)

\({\longrightarrow{\sf{a = \dfrac{27}{180}}}}\)

\({\longrightarrow{\sf{a \approx 0.15}}}\)

\(\star{\underline{\boxed{\tt{\red{a \approx 0.15 \: m/s^{2} }}}}}\)

Hence, the acceleration of dog is 0.15 m/s².

\(\rule{300}{1.5}\)

Answer:

The pressure at the bottom of the river is less than that at the bottom of the lake.

Explanation:

From Bernoulli's equation, the pressure difference is given by

ΔP = ρgΔh + ρ(v₂² - v₁²)/2 where ρ = density of water, g = acceleration due to gravity, Δh = depth, v₁ = velocity at top, v₂ = velocity at bottom

For the lake, v₁ = v₂, since the velocity at the top and bottom are the same. So,

ΔP₁ = ρgΔh + ρ(v₁² - v₁²)/2 = ρgΔh + 0 = ρgΔh

P₂ - P₁ = ρgΔh

P₂ = P₁ + ρgΔh

For the river, v₁ < v₂, since the velocity at the top of the river is greater than at the bottom.

So,

ΔP₂ = ρgΔh + ρ(v₂² - v₁²)/2.

Since  v₁ < v₂, ρ(v₂² - v₁²)/2 will be negative,

So,

ΔP₂ = ρgΔh - ρ(v₂² - v₁²)/2.

Since ρ(v₂² - v₁²)/2 is negative, making ΔP less than that in the lake.

So, ΔP₂ = ΔP₁ -  ρ(v₂² - v₁²)/2.

ΔP₂ = P₃ - P₁

P₃ - P₁ = P₂ - P₁ -  ρ(v₂² - v₁²)/2.

P₃ = P₂ - ρ(v₂² - v₁²)/2.

where P₃ = pressure at bottom of the river and P₂ = pressure at bottom of the lake and P₁ = atmospheric pressure at top of river and lake respectively.

Since the factor ρ(v₂² - v₁²)/2 is removed from the pressure at the bottom of the lake, the pressure at the bottom of the river is therefore less than that at the bottom of the lake.

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Magnetic field are a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are produced by moving electric charges. Everything is made up of atoms, and each atom has a nucleus made of neutrons and protons with electrons that orbit around the nucleus. Since the orbiting electrons are tiny moving charges, a small magnetic field is created around each atom.

Similarities between magnetic fields and electric fields: Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like pole repel unlike poles attract. Electric field points in the direction of the force experienced by a positive charge.

0.09 Nm is net torque about the axle

F=9 N

r=10 cm= 0.01 m

Torque= rF

Torque=9×0.01

Torque=0.09 Nm

In physics, torque, which is also known as the moment of a force, is the propensity of a force to rotate the body to which it is applied. According to the rotational axis, the torque is determined by multiplying the force component's magnitude in the plane perpendicular to the axis by the shortest distance between the axis and the force component's direction. The force vector may always be located in a plane parallel to the axis, regardless of its orientation in space. In SI units, torque is expressed in newton meters.

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1. The deceleration of the aeroplane in the 35s is -1.6m/s²

2. The force acting on the aeroplane is 4.0× 10⁵N

3. The momentum of the aeroplane when its speed is 6.0m/s is 15 ×10⁶kgm/s

The equation of motion are used in solving problems related to motion. The equations of motion are

1. v = u+ at

2 S = ut + 1/2at²

3. v² = u²+2as

where v is the final velocity

u is the initial velocity

S is the distance

t is time

a is the acceleration

The deceleration of the plane after 35s can be calculated as;

v = u+at

v= 6m/s, u= 62m/s , t = 35s

6 = 62+35a

35a = 6-62

35a =- 56

a = - 56/35

a = -1.6m/s²

the negative sign shows that the plane decelerates.

The force acting on the plane is calculated as;

F = ma

F = 2.5×10⁵× 1.6

F = 4× 10⁵N

The momentum of the plane at 6m/s is ;

p = mv

p = 2.5×10⁵ × 6

p = 1.5× 10⁶ kgm/s

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The net force F4 exerted on the runner in lane 4 will be 2 times the net force F1 exerted on the runner in lane 1.

What is newton's second law?

Newton's second law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

This claim is correct because Newton's Second Law states that the net force on an object is equal to the mass times the acceleration of the object.

Therefore, since the runners have the same mass, the acceleration of the runner in lane 4 must be twice that of the runner in lane 1 to account for the difference in speed.

Since acceleration is directly proportional to the net force on an object, this implies that the net force F4 exerted on the runner in lane 4 is twice the net force F1 exerted on the runner in lane 1.

Therefore, The net force F4 exerted on the runner in lane 4 will be 2 times the net force F1 exerted on the runner in lane 1.

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The position of the rock after 3 seconds can be calculated using the equation for the vertical motion of a projectile. The velocity of the rock after 3 seconds can also be determined by taking into account the effects of gravity.

To calculate the position of the rock after 3 seconds, we can use the equation for vertical displacement in projectile motion:

h = h0 + v0t - (1/2)g\(t^2\)

where h is the final position, h0 is the initial position (which is 0 in this case, as the rock is thrown from the edge of the cliff), v0 is the initial velocity (which is 13 m/s upwards), t is the time (3 seconds), and g is the acceleration due to gravity (approximately 9.8 m/\(s^2\)).

Plugging in the values, we have:

h = 0 + (13 m/s)(3 s) - (1/2)(9.8 m/\(s^2\))(3 s\()^2\)

h = 0 + 39 m - (1/2)(9.8 m/\(s^2\))(9 \(s^2\))

h = 0 + 39 m - 44.1 m

h ≈ -5.1 m

Therefore, the position of the rock after 3 seconds is approximately 5.1 meters below the initial position at the edge of the cliff.

To calculate the velocity of the rock after 3 seconds, we can use the equation for vertical velocity in projectile motion:

v = v0 - gt

where v is the final velocity, v0 is the initial velocity (which is 13 m/s upwards), t is the time (3 seconds), and g is the acceleration due to gravity (approximately 9.8 m/\(s^2\)).

Plugging in the values, we have:

v = (13 m/s) - (9.8 m/\(s^2\))(3 s)

v = 13 m/s - 29.4 m/s

v ≈ -16.4 m/s

Therefore, the velocity of the rock after 3 seconds is approximately 16.4 m/s downwards.

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pertaining to the oscilloscope and in accordance with the details in the laboratory manual. what axis quantifies the voltage intensity? Voltage (on the vertical axis) versus time is plotted on the oscilloscope's graph (on the horizontal axis).

An oscilloscope is a class of electronic test equipment that graphically shows changing electrical voltages as a two-dimensional plot of one or more signals as a function of time.

The major goals are to show on the screen waveforms that are repeating or singular and would otherwise occur too quickly for the human eye to see. The waveform that is being presented can then be examined for characteristics like amplitude, frequency, rising time, time interval, distortion, and others.

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Work done on the jet by the catapult is 3.31x10⁷j

We are given that,

Thrust = F = 2.10105N

As a result, the work performed by the engines and the catapult will be equal to the kinetic energy of the jet at takeoff. then the formula can be used to determine the work the jet did.

Work done by the engines = Force x Distance

Work done by the engines,

(2.10 X 10⁵J)x 90m = 1.89 x 10⁷J

Total kinetic energy = 5.20x 10⁷J

The total kinetic energy minus the work produced by the engines is the amount of work the catapult does on the jet.

Therefore the

work done on the jet by the catapult =( 5.20 X 10⁷J) -(1.89 X 10⁷) work done on the jet by the catapult = 3.31x10⁷j

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fjfjfj

Explanation:

rufifiriirifieif

Answer:it’s all of the above for the ck-12

Explanation:

The constant for the permeability of the of the medium (air) is. This number can be reduced to 1.256x10⁻⁶ tm/a. Option b is correct.

The constant for the permeability of air is typically represented by the symbol "μ" and has a value of approximately 4π x 10⁻⁷ T·m/A. This value is also known as the "permeability of free space" or "vacuum permeability". While the value cannot be reduced to zero, it can be reduced to a lower value in certain materials.

For example, the permeability of ferromagnetic materials can be much higher than the permeability of air, while non-magnetic materials may have a permeability that is very close to that of free space. However, none of the options provided correspond to these values. The closest option is b. 1.256x10⁻⁶ tm/a, which is the permeability of free space divided by 4π. Hence Option b is correct.

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Answer:

Festival dances should be introduced to the younger generations of today because they teach them about different cultures. Festival dances are also a way to be active and have a fun time.

Explanation:

The horizontal component of the velocity is equal to: D. 15 m/s.

Given the following data:

To determine the horizontal component of the velocity:

The horizontal component of the velocity represents the influence of velocity  in displacing an object or projectile in the horizontal direction.

Mathematically, the horizontal component of velocity is given by the formula:

\(V_x = Vcos(\theta)\)

Substituting the given parameters into the formula, we have;

\(\\\\V_x = 30cos(60)\\\\V_x = 30 \times 0.5\)

Horizontal component, Vx = 15 m/s

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The force F1 is equal and opposite to the downward force thus, F1 is equal to 60 N. The force F2 is inclined to 30 ° from leftward force and it is equal to 38.97 N in magnitude.

Force is an external agent acting on a body to deform it or to change its state of motion or rest. Force is a vector quantity and it is characterised by its magnitude and direction.

If two forces acting on a body from the same directions, then the net force will be the sum of these two forces. If they are acting from opposite directions, they will cancel each other in magnitude.

The force F1 is equal and opposite to the force acting downward. Thus its magnitude is 60 N. The force F2 is inclined to 30 ° from horizontal direction.

F2 = 45 cos 30 = 38.9 N.

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Answer:

Electrons

Explanation:

Ionised atoms loose and gain electrons.

hope it helps!

Answer:

a

Explanation:

because there both in the solar system

There are billions of galaxies in the universe and each contains trillions of stars which are different in size, distance from the earth, brightness etc.When we scientifically investigate about this we can obtain the stars of similarity to sun in brightness and size.

Galaxies are a group of stars, dust and dark matter joined together by the force of gravitation. Three trillions of stars in each galaxies. Some stars forms a group of certain  size and shape and number also they are called constellations.

Sun is the closest star to earth. Thus sunlight is the major source of energy in earth. Sunlight reaches the earth surface within a time of 8 minute and 6 seconds.

The star which nearest to sun is named proximacentory which far from the earth and we can see it as spot of light in the sky. There are many more stars which are giants are more brighter than sun.

The brightness and luminosity is determined based on the distance from the earth. Which can be calculated in terms of light years. Each star has an age and length of life. After that the star may die still their light energy can be seen in sky because their light takes more years to reach here.

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Antonie van Leeuwenhoek was the first to see single-celled organisms. Leeuwenhoek's observations of microorganisms and the development of his own simple microscope, which he used to observe and examine microbial life forms, are significant in the history of microscopy.

His work showed that the microscope was a valuable tool for scientific discovery. Leeuwenhoek's work also established the importance of microorganisms in life processes.A cheek cell is a type of cell that can be seen in human mouths. They appear to be rectangular in shape and have a nucleus in the center. A skin cell is a kind of cell that makes up human skin. It is a type of epithelial cell that is flat and has a nucleus in the center.Both cheek cells and skin cells, on the other hand, are two types of cells that can be seen with a light microscope. Cheek cells and skin cells are much bigger than bacteria, but they are much smaller than the objects Leeuwenhoek saw with his microscope. Leeuwenhoek's discoveries led to the realization that life existed on a small scale, revealing the complexity of even the tiniest forms of life on the planet.

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Answer:

im saying 4. its just a guess it makes sense tho

Explanation:

sorry for getting this wrong.

Answer:

longitudinal waves

Explanation:

I got it right

The maximum angle the ball can travel without leaving the circular track is approximately 31.7 degrees.

When the spring is unstretched, the potential energy stored in it is zero. If the spring is pulled back by 100 mm and released, it will accelerate the 0.3 kg ball.

Since the track is circular, the ball will travel in a circular path. The force acting on the ball will be the tension in the string, which is equal to the force provided by the spring.

Using Hooke's Law, the force provided by the spring is given by F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.

Therefore, the force provided by the spring when it is stretched by 100 mm is F = -(1500 N/m)(0.1 m) = -150 N.

The maximum angle the ball can travel without leaving the circular track can be found by equating the centripetal force to the weight of the ball, which is given by mv^2/r = mg.

Solving for the angle, we get θ = sin^(-1)(g*r/v^2).

To find v, we can use the conservation of energy principle, which states that the initial potential energy stored in the spring is converted to kinetic energy when the spring is released.

Therefore, 1/2*k*x^2 = 1/2*m*v^2, which gives v = sqrt(k/m)*x = sqrt(1500 N/m/0.3 kg)*0.1 m = 7.75 m/s.

Substituting the values, we get θ = sin^(-1)(9.8 m/s^2*0.15 m/7.75 m/s)^2 = 31.7 degrees.

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The maximum angle the 0.3 kg ball will travel without leaving the circular track is approximately 36.87 degrees.

To determine the maximum angle the 0.3 kg ball will travel without leaving the circular track, we can analyze the energy conservation in the system.

Given:

Spring constant (k) = 1500 N/m

The maximum displacement of the spring (s) = 100 mm = 0.1 m

Mass of the ball (m) = 0.3 kg

We'll consider the potential energy stored in the spring when it is compressed and the potential energy of the ball when it is at the maximum angle.

At the initial position, all the energy is stored in the spring:

The potential energy stored in the spring \((Us) = (1/2) * k * s^2.\)

Substituting the values, we find:

Us = (1/2) * 1500 N/m * (0.1 m)^2.

Calculating this expression, we find:

Us = 7.5 J.

At the maximum angle, all the potential energy is converted into the gravitational potential energy of the ball:

Potential energy of the ball (Ug) = m * g * h,

where g is the acceleration due to gravity and h is the height.

Since the ball is on a circular track, the maximum angle is when the ball is at the highest point of the circular track, so h is the radius of the circular track.

The gravitational potential energy can be expressed as:

Ug = m * g * r.

The ball will leave the circular track when the gravitational force equals the maximum centripetal force:

\(m * g = m * v^2 / r,\)

where v is the velocity of the ball.

Simplifying, we find:

\(v^2 = g * r.\)

Since the energy is conserved, we can equate the potential energy of the spring to the potential energy of the ball:

Us = Ug.

Substituting the values, we have:

\((1/2) * 1500 N/m * (0.1 m)^2 = 0.3 kg * g * r.\)

Simplifying, we find:

g * r = 25 m.

Substituting the expression for \(v^2\), we have:

\(v^2 = 25 m.\)

Taking the square root, we find:

v ≈ 5 m/s.

Now, we can calculate the maximum angle using the velocity and the radius:

tan(θ) = v / sqrt(g * r).

Substituting the values, we find:

tan(θ) = 5 m/s / \(sqrt(9.8 m/s^2 * 25 m)\).

Calculating this expression, we find:

tan(θ) ≈ 0.721.

Taking the inverse tangent, we find:

θ ≈ 36.87 degrees (rounded to two decimal places).

Therefore, the maximum angle the 0.3 kg ball will travel without leaving the circular track is approximately 36.87 degrees.

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The force applied to the small piston in order to lift the engine would be 376.56 N.

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

As given in the problem,

the cross-sectional area of the small piston =0.045 m2

the cross-sectional area of the large piston =0.239 m2

weight of the engine = 2000 N

By using the Pascal law,

Force /0.045 = 2000/0.239

Force = 376.56 N

Thus, the force applied on the small piston would be 376.56 N

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Answer:

Radio waves are electromagnetic waves

Explanation:

•Radio waves are electromagnetic waves.

•Satellites are located within our atmosphere.

•Radio waves are very high energy mechanical waves.

•Radio waves are changed to gamma rays for transmission.

Answer:

B. 44.1

Explanation:

We can use the equation y=1/2gt^2 to find the distance in the vertical direction, where y= distance in the vertical direction, g=acceleration due to gravity, and t=time

we only know acceleration due to gravity(9.8m/s^2),

plug in known variables,

y=1/2*9.8*t^2

we still need to plug in the time variable.

In order to do that, we an use the s=d/t equation and re arrange that formula to get: t=d/s, where t=time, d=distance, and s=speed.

plug in known variables,

t=60/20

t=3

now that we know the time, we can plug that into our original equation

y=1/2*9.8*3^2

simplify,

y=4.9*3^2

y=4.9*9

y=44.1m

Thus, B is your answer

The wave speed will be constant. the wavelength will decrease in (a) and the frequency will decrease in (b)

The frequency of a wave can be defined as the number of complete revolution per second made by a vibrating object.

a. When the frequency of a wave increases, what will happen to the wavelength of the wave depends on the medium. But in most cases, the wavelength of the wave will decrease since the wave speed will be constant.

b. When the wavelength of a wave increases, what will  happen to the frequency of the wave is the fact that the wave frequency will be low. Because the product of frequency and wavelength must be constant.

Therefore, when the frequency of a wave increases, the wavelength of the wave will decrease and when the wavelength of a wave increases, the wave frequency will decrease.

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Answer:

B - Average Kinetic Energy decreases

Explanation:

As the water vapor cools the molecules becomes less excited. Less excitement means less movement, hence less kinetic energy

The strength of the second magnetic field is approximately 78.18 mT. The Hall voltage is directly proportional to the magnetic field strength.

Let's recalculate the strength of the second magnetic field using the given information.

Given:

Hall voltage in the first magnetic field, VH1 = 1.9 mV

Hall voltage in the second magnetic field, VH2 = 2.8 mV

Magnetic field strength in the first field, B1 = 55 mT (convert to Tesla: 55 × 10⁻³ T)

We can set up a proportion using the relationship between the Hall voltages and magnetic field strengths:

VH1 / B1 = VH2 / B2

To solve for B2, rearrange the equation:

B2 = (VH2 * B1) / VH1

Substitute the given values:

B2 = (2.8 mV * 55 × 10⁻³) T) / (1.9 mV)

Now, let's convert the millivolt (mV) to volts (V):

B2 = (2.8 × 10⁻³) V * 55 ×  10⁻³) T) / (1.9 ×  10⁻³ V)

Simplifying the expression:

B2 ≈ 78.18 mT

Therefore, the strength of the second magnetic field is approximately 78.18 mT.

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The closest object that can be photographed is 81.63mm and the magnification of this closest object is -0.404.

The focal length of a lens is determined when the lens is focused at infinity. It is obtained from the reciprocal of objects' distance and image distance. Magnification is the enlarged image that is formed over the object size.

From the given,

focal length (f) = 23.5mm

object's distance (u) = 33mm

imagen distance(v) =?

Focal length, (1/f) = 1/u + 1/v

1/v = 1/f - 1/u

    =1/23.5 - 1/33

1/v  = 12.2mm

v = 1/12.2 mm

  = 81.96mm

Thud, the image distance is v= 81.96mm.

Magnification (M) = -v/u

M = -33 / 81.96

  = - 0.402.

Thus, the magnification is -0.402.

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Answer: 1.5 × 10^-17

Explanation:

Given the following :

Frequency(f) = 2.2 × 10^16 Hz

Planck's constant(h) = 6.63 × 10^-34

The energy of a photon 'E' is given as the product of frequency and the planck's constant

E = hf

E = (6.63 × 10^-34) × (2.2 × 10^16)

E = 6.63 × 2.2 × 10^(-34 +16)

E = 14.586 × 10^-18

E = 1.4586 × 10^-17

E = 1.5 × 10-17 (2 S. F)

Answer:

C. 1.5 × 10–16 J

Explanation: