Answer:
Light slows down when it moves from air into water.
Explanation:
It is the property of light that travels faster in a less dense medium.
In a more dense medium, the speed of the light slows down and bends towards the normal.
The air is less dense medium and water is a more dense medium.
When light passes from air to water, the light bends.
This is known as the refraction of light.
Answer:
C is the correct answer on Ed gen uity
Timmy and Sarah go for a hike. they hike 5 miles in three hours and 15 minutes. what is the average speed to the nearest 10th of a mile. (Hint - you need to convert minutes to hours)
The average speed is 1.54 miles per hour.
As we know that,
Average speed [tex]=\frac{Distance}{time}[/tex]
Given that Timmy and Sarah travel 5 miles in three hours and 15 minutes.
Time [tex]=3+\frac{15}{60} =\frac{13}{4}hours[/tex]
Average speed [tex]=5\div \frac{13}{4} =5*\frac{4}{13}=1.54 miles/hour[/tex]
Thus, The average speed is 1.54 miles per hour.
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A 1400 kg vehicle travelling at 22 m/s slows at a constant rate to 4.5 m/s in 6.75 s. What was the net force acting on the car during this time?
Answer:
Net force = 3640 N
Explanation:
From first equation of motion
V = U + at
4.5 = 22 + 6.75a
( 4.5-22) ÷ 6.75 = a
acceleration (a) = -2.6m/s^2 ( negative sign shows deceleration, but actual acceleration is 2.6)
But Force = mass × acceleration
Force = 1400 × 2.6
Force = 3640 N
A 240-volt, 2-amp motor is connected to a three-wire, 120/240-volt system. Connected between the black wire and neutral are four 200-watt, 120-volt lamps and a 120-volt, 1-amp motor. Between the red wire and neutral are three 200-watt, 120-volt lamps, one 1.67-amp motor and one 120-volt, 1-amp motor. (Round the FINAL answer to two decimal places.). How many amps flow in the red wire?
Answer:
(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is 15.36 A (iv) when 240 volt were disconnected current in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A
Explanation:
Solution
The current drawn by one amp is
I =P/V
I =200/120
I= 1.67 A
(i) The current flow in the black wire is
IBK = 4 * 1.67 A + 1A + 2A
IBK = 9.67 A
(ii) Current flow in the red wire is
IRD = 3 * 1.67 A + 1.67 A + 1A + 2A
= 8.68A + 1 A = 9.68 A
Note: Kindly find an attached copy of part of the solution to the given question above.
Which describes a relationship when calculating the energy of a photon? The energy of the photon is directly proportional to frequency. The energy of the photon is inversely proportional to frequency. As the energy of the photon increases, Planck’s constant increases. As the energy of the photon increases, Planck’s constant decreases.
Answer:
The energy of a photon is directly proportional to it's frequency
Answer:
B. The energy of the photon is directly proportional to frequency.
Explanation:
Rank these objects on the basis of their wavelength. (largest to smallest)person v=4.5m/selectron v=0.01cred light v=cproton v=0.01ccar v=27m/sbaseball v=41m/s
Answer:
red light > electron > proton > baseball > person > car
Explanation:
To calculate the wavelength associated to each object, you use the Broglie's relation:
[tex]\lambda=\frac{h}{mv}\\\\[/tex]
h: Planck,s constant = 6.62*10^-34 Js
m: mass
v: velocity
For each object you use an average values of its mass.
person:
mass = 80kg
[tex]\lambda=\frac{6.62*10^{-34}Js}{(80kg)(4.5m/s)}=1.83*10^{-36}m[/tex]
electron:
mass = 9.1*10^{-31}kg
[tex]\lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(0.01c)}\\\\\lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(0.01(3*10^{8}m/s))}\\\\\lambda=2.42*10^{-10}m[/tex]
red light:
In this case you use the following formula:
[tex]\lambda=\frac{c}{f}=\frac{3*10^8m/s}{4.3*10^{14}Hz}=6.97*10^{-7}m\approx700nm[/tex]
proton:
mass = 1.67*10^{-27}kg
[tex]\lambda=\frac{6.62*10^{-34}Js}{(1.67*10^{-27}kg)(0.01(3*10^8m/s))}\\\\\lambda=1.32*10^{-13}m[/tex]
car:
mass = 1500kg
[tex]\lambda=\frac{6.62*10^{-34}Js}{(1500kg)(27m/s)}=1.63*10^{-38}m[/tex]
baseball:
mass = 0.145kg
[tex]\lambda=\frac{6.62*10^{-34}Js}{(0.145kg)(41m/s)}\\\\\lambda=1.11*10^{-34}m[/tex]
hence, by comparing the wavelengths of the objects you have:
red light > electron > proton > baseball > person > car
Which planets have craters?
Answer:Mercury, Venus, Earth and Mars.
Explanation: The surfaces of asteroids and the rocky, ice covered moons of the outer gas planets are cratered as well.
A regular, repeating arrangement of atoms, ions, or molecules is
This a physics problem please solve the elastic collision problem use guess if you can if you can’t it’s fine
I will use my guess because I can. Unless you're the same person just spamming their question but if not, don't worry about it. they will crash at 5 m/s.
Final answer:
To analyze an elastic collision, one should use the momentum and kinetic energy conservation equations, simplifying them if one object is initially at rest. Inelastic collisions involve the conservation of momentum only and require experimental measures to determine energy dissipation.
Explanation:
An elastic collision is a type of collision where both momentum and kinetic energy are conserved. To solve a problem involving an elastic collision, one must use two independent equations: the conservation of momentum equation and the conservation of kinetic energy equation. If one of the objects is initially at rest, as stated in the question, the equations become simpler. We can set the initial velocity of the second object to zero, and then combine the equations algebraically to solve for the final velocities of both objects.
In the case of an inelastic collision, only momentum is conserved, and some kinetic energy is transformed into other forms of energy, such as heat or sound. To analyze changes in kinetic energy for an inelastic collision, one could modify an experiment to measure velocities before and after the collision and then calculate the kinetic energy to determine the fraction of mechanical energy dissipated.
In summary, analyzing an elastic collision typically involves mathematical calculations of velocities and kinetic energy, while an inelastic collision would be explored through experiments and calculations focusing on energy dissipation. If making graphical representations, one would plot momentum and kinetic energy before and after the collision to verify these relationships.
How much force is required to produce a torque of 100 Nm if the force is applied at an angle of 40 at a distance of 25 cm from the pivot?
Answer:
Force, F = 622.28 N
Explanation:
It is given that,
Torque, [tex]\tau=100\ N-m[/tex]
Angle between force and displacement, [tex]\theta=40^{\circ}[/tex]
Distance, d = 25 cm = 0.25 m
We need to find the force required to produce the torque. Torque produced by an object is given by :
[tex]\tau=Fd\sin\theta[/tex]
F is force required to produce torque
[tex]F=\dfrac{\tau}{d\sin\theta}\\\\F=\dfrac{100}{0.25\times \sin(40)}\\\\F=622.28\ N[/tex]
So, applied force is 622.28 N.
Why is it a good idea to discuss gravity and fluid flows when learning about electricity?
Answer:
The flow of fluid molecules in fluid mechanics is synonymous to the flow of electrons in electricity.
Explanation:
The flow of fluid molecules in fluid mechanics is synonymous to the flow of electrons in electricity. Water, when it is falling by the force of gravity, can be used to turn turbines and generators that produce electricity.
A turbine converts the kinetic energy of falling water into mechanical energy. Then a generator converts the mechanical energy from the turbine into electrical energy. Hydroelectric power plants are the most efficient means of producing electric energy.
Discussing gravity and fluid flows is beneficial when learning about electricity due to the analogies between these phenomena, which aid in understanding the behavior of electric charges and circuits. These comparisons help visualize electrical concepts and deepen comprehension of various electrical phenomena and technologies.
Discussing gravity and fluid flows when learning about electricity is crucial due to the similarities in the concepts governing all these phenomena. For instance, electricity is often taught through the analogy of water flow in pipes, where electric current, voltage, and resistance are likened to the flow rate, pressure difference, and pipe size in fluid dynamics. This analogy helps students visualize and understand how electric charges behave in circuits, just as water flows through a system under the influence of gravity. Moreover, the concept of gravitational potential energy in physics is analogous to electric potential energy in electromagnetism, where masses and charges experience forces due to their respective fields. Understanding these analogies deeply enriches the comprehension of electrical phenomena and leads to insights into the behavior of electricity in various mediums, such as solids, liquids, and gases.
Examples such as lightning, the formation of bluish crust on car battery terminals, and the need for electrolytes in sports drinks can be understood better by studying the flow of charged particles. These particles, like fluid in the presence of a gravitational field, respond to electric fields by moving from areas of high potential to low potential, analogous to water flowing downward due to gravity. This comparison between electric and gravitational fields lays a solid foundation for exploring more complex electrical behaviors and technologies.
3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming to rest. How much work was done on the puck?
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by
[tex]W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J[/tex]
The work done on the puck is 12.3 J.
What is the total mechanical energy of a 200 kg roller coaster moving with a velocity of 16 m/s at a height of 18 m above the ground?
Mechanical Energy = PE + KE
PE: mgh = 200 x 9.8 x 18 = 35280
PE: 35280 Joules
KE: 1/2mv^2 = 1/2 x 200 x 16^2 = 25600
KE: 25600 Joules
ME: 35280 + 25600
ME: 60,880J
An open organ pipe is 2.46 m long, and the speed of the air in the pipe is 345 m/s.
a. What is the fundamental frequency of this pipe?
Answer:
Fundamental frequency is 70.12 m
Explanation:
For an open organ pipe, the fundamental frequency is given by :
[tex]f=\dfrac{nv}{2l}[/tex]
n = 1 for fundamental frequency
v is speed of sound in air, v = 345 m/s
l is length of open organ pipe, l = 2.46 m
Substituting values in above formula. So,
[tex]f=\dfrac{1\times 345}{2\times 2.46}\\\\f=70.12\ Hz[/tex]
So, the fundamental frequency of this pipe is 70.12 m.
Final answer:
The fundamental frequency of a 2.46 m long open organ pipe, with air speed 345 m/s, is approximately 70.12 Hz.
Explanation:
To find the fundamental frequency of an open organ pipe, we can use the formula for the fundamental frequency of a pipe that is open at both ends, which is f1 = v / (2*L), where f1 is the fundamental frequency, v is the speed of sound in the pipe, and L is the length of the pipe.
In this case, the organ pipe is 2.46 meters long, and the speed of the air in the pipe is given as 345 meters per second. Plugging these values into the formula, we get:
f1 = 345 m/s / (2 * 2.46 m) = 345 m/s / 4.92 m = 70.12 Hz
Therefore, the fundamental frequency of this 2.46 m long open organ pipe is approximately 70.12 Hz.
A 0.5 kg ball is at the top of a ramp which is 8 meters high. How much
Vinetic energy does the ball have at the bottom of the ramp? (round the
answer to the tenths place)
Answer:
same as the potential energy possessed at top
= 0.5×9.8×8
= 39.2 joules
Answer: The answer is 40 J.
Explanation: I did it just now.
as the angle of incidence of light increases, what happens to the angle of refraction?
Answer:
it increases too as it bend away from the normal line
A wave with a frequency of 60.0 Hz travels through rubber with a wavelength of .90 m. What is the speed of this wave?
Answer:
Speed of wave 54 ms⁻¹.
Explanation:
Given data:
Frequency of wave = 60 Hz
Wavelength of wave = 0.90 m
Speed = ?
Solution:
Formula
speed = wavelength × frequency
Now we will put the values in formula.
v = f × λ
Hz = s⁻¹
v = 60 s⁻¹ × 0.90 m
v = 54 m s⁻¹
A wave with a frequency of 60 Hz and a wavelength of 0.90 m through rubber travels at a speed of 54 m/s.
Explanation:The speed of a wave can be calculated by the formula: Speed = Frequency * Wavelength. Given that the frequency is 60.0 Hz and the wavelength is 0.90 m, we can substitute these values into the formula. Therefore, the speed of the wave is 60.0 Hz * 0.90 m = 54 m/s. This means that a wave with a frequency of 60 Hz and a wavelength of 0.90 m travels through rubber at a speed of 54 m/s.
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In which direction is the magnetic force acting in the charge?
Answer:
"Right Hand
Right Hand Rule: Magnetic fields exert forces on moving charges. This force is one of the most basic known. The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1) "
Explanation:
Answer:
The direction of the magnetic force acting on a moving electric charge in a magnetic field is perpendicular to the direction of motion. A magnetic force is exerted on an electric charge moving through a uniform magnetic field. An electric charge moving parallel to a magnetic field experiences a magnetic force.
Explanation:
An object with a mass of 7 kg accelerates 5 m/s2 when an unknown force is applied to it. What is the unknown force?
Answer:
35 N
Explanation:
Force, F = ma where m = mass = 7 kg and a = acceleration = 5 m/s².
So, F = ma = 7 kg × 5 m/s² = 35 N
Final answer:
To find the unknown force applied to an object with a mass of 7 kg accelerating at 5 m/s², use the formula F = ma to calculate it as 35 N.
Explanation:
The unknown force applied to the object with a mass of 7 kg can be calculated using the formula F = ma.
Given mass (m) = 7 kg and acceleration (a) = 5 m/s²
Substitute the values into the formula: F = (7 kg) (5 m/s²) = 35 N
Therefore, the unknown force applied to the object is 35 Newtons.
Figure 15-3 In Figure 15-3, which gas makes up the greatest percentage of Earth's atmosphere?
Answer:
Nitrogen
Explanation:
A man is 6ft tall. What is the smallest size plane mirror he can use to see his entire image.
The smallest size plane mirror he can use is 3 feet tall, ensuring he can see his entire image reflected in the mirror
To see his entire image in a plane mirror, the mirror needs to be at least as tall as the man's height. However, since only half of his height is required for the mirror, the smallest size plane mirror he can use is half his height.
Given that the man is 6 feet tall, the smallest size plane mirror he can use to see his entire image would be [tex]\( \frac{6}{2} = 3 \)[/tex] feet tall. Therefore, the smallest size plane mirror he can use is 3 feet tall.
How much power can a motor output if it does 25 J of work in 1 s? *
Answer:
25 W
Explanation:
Power: This can be defined as the rate at which work is done. The S.I unit of power is Watt (W).
The expression of power is given as
Power = work output/time
P = W/t................ Equation 1
Given: W = 25 J. t = 1s
Substitute into equation 1
P = 25/1
P = 25 W
Hence the output power of the motor is 25 W
Answer:
The power is 25 W
Explanation:
Given;
work done by motor output. E = 25 J
Time taken to complete this task, t = 1 s
Power is the amount of work done per unit time or energy per unit time.
The amount of power developed by the motor, can be calculated as follows using the formula below;
[tex]Power = \frac{Energy}{time} = \frac{25}{1} = 25 \ W[/tex]
Therefore, the amount of power developed by the motor in the given time is 25 W.
How do spectrographs help astronomers classify stars?
They determine the size of stars.
They evaluate the color of stars.
They measure the brightness of stars.
They analyze the composition of stars.
Spectrographs help astronomers classify stars by providing a 'fingerprint' of the star. This fingerprint allows astronomers to determine the composition, temperature and other factors of a star.
Explanation:Spectrographs are critical tools in astronomy because they allow astronomers to analyze and classify stars based on their spectra. When light from a star passes through a spectrograph, it is split into its component colors, creating a spectrum. This spectrum is like a fingerprint for the star because each element leaves a specific pattern of spectral lines in the star's light. In other words, by identifying these lines, astronomers can determine the composition of stars.
In addition to determining a star's composition, the spectrum can also give hints about a star's temperature, velocity relative to Earth, and even information about its size. So, while a spectrograph does not directly measure the size or brightness of stars, the data it provides enables astronomers to infer this information.
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The correct answer is option D. They analyze the composition of stars.
A spectrograph helps astronomers classify stars by analyzing their spectra to determine chemical composition, temperature, and atmospheric pressure. This information differentiates between various types of stars. Stellar spectra also yield data on a star's motion and rotation.A spectrograph helps astronomers classify stars by analyzing the composition of stars through their spectra. By measuring the spectrum of a star, astronomers can determine its chemical composition, temperature, and atmospheric pressure. This information is vital for distinguishing between different types of stars, such as giant stars with large radii and low atmospheric pressures compared to smaller stars with high atmospheric pressures.The differences in the spectra of stars are mainly due to their temperatures, which are categorized into spectral classes. Each element in a star emits or absorbs light at specific wavelengths, allowing astronomers to identify the elements present in the star. Additionally, stellar spectra provide data on a star's motion (radial velocity) and rotational speed through the Doppler effect.Two people push on the same door from opposite sides as shown. A man pushes on a door from the left. A woman pushes on the door from the right. They will only see the door move when both people exert the same force so that the forces are unbalanced. neither person exerts a force on the door so that the net force is zero. one person exerts more force than the other so that the forces are unbalanced. both people exert the same force so that the forces are balanced.
Answer:
one person exerts more force than the other so that the forces are unbalanced.
Explanation:
Answer:
C.one person exerts more force than the other so that the forces are unbalanced
Explanation:
just took the test
(ii)
Calculate the amount of energy required to melt 15 kg of ice at 0 °C.
Specific latent heat of fusion of ice = 3.4 x 105 J/kg.
Energy =
Answer:
uekdjdbrjrnkfkdkfkff
Explanation:
E=mc(tetha)
=15(3.4x105)(0)
=0joules
A red car passes a blue car. Which is true?
A.
The blue car must be parked.
B.
The red car must be moving backwards.
C.
The blue car must be moving slowly.
D.
The red car is moving faster than the blue car.
Answer:
D
Explanation:
Amphibians live part of their lives like:___
A.Mammals
B.Fish
C.Birds
If you drop something from low height is it faster
Faster to hit the ground because if so yes?
You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room whose walls are kept at -5°C, what is the net flow rate of energy out of the sphere?
The net rate of energy flow from the tungsten sphere can be calculated using the modified Stefan-Boltzmann equation, taking into account the sphere's emissivity, surface area and temperature, as well as the temperature of the surroundings. After calculating, we have the net rate of heat transfer.
Explanation:The net flow rate of energy from an object, in this case a tungsten sphere, can be calculated using the modified Stefan-Boltzmann equation which is suitable for net rate of heat transfer by radiation. The formula used is Qnet = σeA(T₁⁴ - T₂⁴), where σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ J/s.m².K⁴), e is the emissivity (0.35 for tungsten), A is the surface area of the sphere (πr² with r being the radius of the sphere), and T₁ and T₂ are the temperatures of the sphere and the surroundings respectively, converted to Kelvin.
First, we convert the Celsius temperatures to Kelvin by adding 273.15 to each temperature, making T₁ = 298.15 K and T₂ = 268.15 K. Then, finding the surface area, we get A = 4πr² = 4.90 m² with r = 0.25 m. Substituting these values into the equation gives us Qnet = (5.67 × 10⁻⁸ J/s.m².K⁴)(0.35)(4.90 m²)[(298.15 K)⁴ - (268.15 K)⁴] which, when calculated, gives the net rate of heat transfer by radiation from the tungsten sphere.
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Final answer:
The net flow rate of energy from a tungsten sphere in a cooler room can be calculated using the Stefan-Boltzmann law; accounting for emissivity, surface area, and the difference in the fourth powers of the absolute temperatures of the sphere and its surroundings.
Explanation:
The student's question regarding the net flow rate of energy out of a tungsten sphere enclosed in a room with cooler walls can be answered using the Stefan-Boltzmann law for thermal radiation which states that the power radiated from an object is proportional to its emissivity (e), the Stefan-Boltzmann constant (σ), the surface area of the object (A), and the fourth power of its absolute temperature (T) in kelvins. For our purposes, the net power radiated is the difference in power radiated by the sphere and the power absorbed from the surroundings. To find the surface area (A) of the sphere we use the formula 4πr2 where r is the radius of the sphere. Then, we transform the temperatures from Celsius to Kelvins and plug our values into the equation P = eσA(T4 - T04) to find the net flow rate of energy out of the sphere.
Two charged parallel plates are 0.25 meters away from each other. The field between the plates is 600 . What is the electric potential difference? ΔV = V
Answer:
ΔV = 150 V
Explanation:
What is potential energy?
Energy released
Energy stored
Moving Energy