Answers
Explanation:
Interference is the example of soap bubble colouringEXTRA INFO:(LOOK AT THE IMAGE)
An incoming light ray is partly reflected by the top surface of the soap film and partly reflected by the bottom surface. The wave reflected from the bottom surface has traveled further (an extra distance equal to twice the thickness of the film) so emerges out of step with the top wave. When the two waves meet, they add together, and some colors are removed by destructive interference. Where the film is thickest, the bubble appears more blueish; where it's thinner, it will look more violet or magenta.
[tex]\huge\bold\color{black}{ANSWER}[/tex]
Soap bubble coloring example: INTERFERENCE
Related Questions
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
Answers
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m
Answers
Answer:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]
So, the sound travel 4.86 m between wings flaps.
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator
Answers
Answer:
f = 347.08 N
Explanation:
The frictional force exerted by the floor on the refrigerator is given as follows:
[tex]f = \mu R = \mu W[/tex]
where,
f = frictional force = ?
μ = coefficient of static friction = 0.58
W = Weight of refrigerator = mg
m = mass of refrigerator = 61 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 347.08 N
3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)
Answers
Answer:
150 N
Explanation:
Given that,
Acceleration (a) = 3 m/s²Mass of the bike (m) = 50 kgWe are asked to calculate force required.
[tex]\longrightarrow[/tex] F = ma
[tex]\longrightarrow[/tex] F = (50 × 3) N
[tex]\longrightarrow[/tex] F = 150 N
Convierta 8.5mW a cal/h (1 cal=4.186 j)
Answers
Answer:
[tex] = { \bf{2.03 \times {10}^{ - 6} }}[/tex]
a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
Answers
Answer:
Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?
This problem has been solved!
This problem has been solved!See the answer
This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?
Explanation:
use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity
In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answers
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answers
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
The following contribute to accidents when a teen driver has other teens as passengers
Answers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
To learn more about factors contribute to accidents refer to:
https://brainly.com/question/4853141
#SPJ2
A 0.495-kg hockey puck, moving east with a speed of 4.50 m/s , has a head-on collision with a 0.720-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision?
Answers
Answer:
a) [tex]v_1=-0.833m/s[/tex]
b) [tex]V_2=12.5m/s[/tex]
Explanation:
From the question we are told that:
Hockey puck Mass [tex]m_1=0.495kg[/tex]
Hockey puck Speed [tex]u_1=4.50m/s[/tex]
Puck Mass [tex]m_2=0.720kg[/tex]
Assuming
Initial speed of Puck [tex]u_2=0[/tex]
Generally the equation for Speed of First Puck is mathematically given by
[tex]v_1=(\frac{m_1-m_2}{m_1+m_2})*v_1+(\frac{2m_2}{m_1+m_2})u_2[/tex]
[tex]v_1=(\frac{0.495-0.720}{0.495+0.720})*4.50+0[/tex]
[tex]v_1=-0.833m/s[/tex]
Generally the equation for Speed of Second Puck is mathematically given by
[tex]V_2=(\frac{2m_1}{m_1+m_2})u_2-(\frac{m_1-m_2}{m_1+m_2})v[/tex]
[tex]V_2=(\frac{2*0.495}{0.495*0.720})*4.50-0[/tex]
[tex]V_2=12.5m/s[/tex]
Why is the force of attraction between the Earth and ourselves so huge compared to the attraction between two apples?
Answers
Answer:
Answer in explanation
Explanation:
The force of attraction between two bodies is governed by Newton's Law of Gravitation:
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
where,
G = Universal Gravitational Constant
m₁ = mass of the first body
m₂ = mass of the second body
r = distance between the two bodies
F = Force
Hence, it is clear from the formula that the magnitude of the force is directly proportional to the product of the masses of the objects. So in the case of the earth and ourselves, the mass of the earth is very large in order of 10²⁴ kg. Due to this huge mass, the attraction between the earth and ourselves is so huge as compared to the attraction between two apples. Because the masses of the apple are very small in grams.
A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.
Calculate:
Area of the largest face
Answers
Answer:
1.5
x 1.0
1.50
x 0.5
075.00
answer: 75.00m
Explanation:
I hope this help
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Answers
Explanation:
Let T is the period of a pendulum. The SI unit of time is seconds (s).
It depends on the acceleration of gravity, g, and the length of the pendulum, l.
The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.
If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.
So,
[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]
Hence, this is the required solution.
the speed of the pulse depends on what?
Answers
Answer:
The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.
Explanation:
Hope this helps.
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answers
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
A flag pole 18m high casts a shadow 9.6m long . Find the distance of top of pole from the far of end of Shadow.
Answers
Answer:
[tex]{ \bf{pythogras \: theorem :}} \\ \\ { \tt{ = \sqrt{ {9.6}^{2} + {18}^{2} } }} \\ = 20.4 \: cm[/tex]
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answers
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
If the force on an object is negative, what is known about the change in velocity?
O The velocity will decrease.
O The velocity will increase.
O The velocity will not change.
O The object will change direction.
Answers
A 5.70 g lead bullet traveling at 490 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet
Answers
Answer:
461.73 K
Explanation:
Given that,
The mass of a bullet, m = 5.7 g
Speed of the bullet, v = 490 m/s
Half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree.
Using the conservation of energy,
[tex]\dfrac{1}{2}(\dfrac{1}{2}mv^2)=mc\Delta T\\\\\Delta T=\dfrac{v^2}{4c}[/tex]
Where
x is the specific heat of lead, c = 130 J/kg K
So,
[tex]\Delta T=\dfrac{(490)^2}{4\times 130}\\\\=461.73\ K[/tex]
So, the increase in temperature of the bullet is 461.73 K.
g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answers
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?
Answers
Answer:
P = 0.14 hp
Explanation:
The power required by the ship is given as:
[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]
where,
P = Power = ?
m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 2 m
t = time = 1 s
Therefore,
[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]
Converting to horsepower (hp):
[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]
P = 0.14 hp
Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answers
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocity due to the wind is 10.0 cm/s, what is the vertical component velocity? Show all work.
Answers
Answer:
24 cm/s
Explanation:
Applying
Pythagoras theorem,
a² = b²+c²............. Equation 1
Where a = resultant, b = vertical component, c = horizontal component
From the question,
Given: a = 26 cm/s, c = 10 cm/s
Substitute these values into equation 1
26² = b²+10²
676 = b²+100
b² = 676-100
b² = 576
b = √576
b = 24 cm/s
A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is
Answers
Answer:
the weight of the air in pound-force (lb-f) is 325.8 lbf
Explanation:
Given;
dimension of the room, = 15 ft by 15 ft by 20 ft
density of air in the room, ρ = 0.0724 lbm/ft³
The volume of air in the room is calculated as;
Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³
The mass of the air is calculated as;
mass = density x volume
mass = 0.0724 lbm/ft³ x 4,500 ft³
mass = 325.8 lb-m
The weight of the air is calculated as;
Weight = mass x gravity
Weight = 325.8 lb-m x 32.174 ft/s²
Weight = 10482.29 lbm.ft/s²
The weight of the air in pound-force (lb-f) is calculated as;
1 lbf = 32.174 lbm.ft/s²
[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]
Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answers
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration
Answers
Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 meters?
Answers
Answer: Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
Explanation:
To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.
This wave base is equal to half of the wavelength. The equation becomes:
Wave base = [tex]\frac{\text{Wavelength}}{2}[/tex]
We are given:
Wavelength = 24 m
Putting values in above equation, we get:
Wave base = [tex]\frac{24m}{2}=12m[/tex]
Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
Drag the titles to the correct boxes to complete the pairs.
Answers
7. If a load of 300N is pulled along the inclined plane shown in the figure, answer the following. B 200 N 0.5m 2m 300 N А i. Calculate the VR and MA of the inclined plane. Calculate the input work and output work. ii.what efficiency of inclined plane?iv.what should be the length of inclined plane if same load has to be pulled with a50N effort for the same efficiency as above
Answers
Explanation:
700n I think friend .. if worng
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answers
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
