Physics College

Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for the function in the figure?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.

Consider The F(x) = Cos(x-C) Function Shown In The Figure In Blue Color, Where 0 C 2. What Is The Value

Answers

Answer 1

The value of parameter C for the function in the figure is 2.

What is amplitude of a wave?

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

A is amplitude of the waveC is phase difference of the wave

What is angular frequency of a wave?

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

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Related Questions

what is elecric field line​

Answers

A vector field and a starting position inside the field define the locus as a "field line." We have electric field lines for the electric fields.

What do you mean by electric field line?

In general, an electric field line is a curve drawn with each point's tangent pointing in the direction of that point's net field. An arrow on the curve is obviously required to designate the direction of the electric field from the two potential directions provided by a tangent to the curve. A field line is a space curve or a three-dimensional curve.The vector gets shorter as you move away from the origin and always points radially outward because the electric field varies as the inverse of the square of the distance that points from the charge.The density of the lines indicates the size of the field. This indicates that the location with a high density of field lines has a stronger electric field due to the charged substance. The electric field is weaker in the area where the density of these lines is low.The electric field can be depicted by joining these vectors to form a line.The direction of an imaginary line called an electric field line at any place must match the direction of the field there.In general, a field line is a curve drawn so that each point's tangent points in the direction of the net field.

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A force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
A). What is the potential energy of the spring when it is stretched 0.700 m?
B). What is its potential energy when it is compressed 8.00 cm?

Answers

Answer:

399j and 456j

Explanation:

(a)spring energy= force * distance

spring energy=570 * 0.700= 399 joules

(b) convert 8.00 cm to m we have 0.08m

E =f * s

E=570 *0.08= 456j

The potential energy of the spring when it is stretched 0.700 m would be  199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.

What is the spring constant?

The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.

The mathematical relation for calculating the spring constant is as follows

F = - Kx

As given in the problem, if a  force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.

k = 570 / 0.7

k = 814.28 N / m

 

The potential energy stored  in spring when it is stretched 0.700 m

E = 1/2 kx²

 =0.5×814.28×0.7²

 = 199.5 Joules

The potential energy stored  in spring when it is stretched 8 cm

E = 1/2 kx²

 =0.5×814.28×0.08²

 =2.60 Joules

Thus, the potential energy of the spring when it is stretched 0.700 m would be  199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.

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A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium filled balloon by a string, holding a significant part of his weight. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond. (the anchor is initially inside the boat)
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.

Answers

(a) The son pops the helium balloon (unchanged)

(b) The fisherman knocks the tackle box overboard and it sinks to the bottom (rises)

(c) The son finds a cup and bails some water out of the bottom of the boat (falls)

(d) The fisherman lowers himself in the water and floats on his back (unchanged)

(e) The fisherman lowers the anchor and it hangs one foot above the bottom of the pond (rises)

(f) The son gets in the water, looses his grip on the string, letting the balloon escape upwards (rises).

Displaced volume of water in the pond

According to Archimedes principle, when a body is partially or completely immersed in a fluid, it experiences an upthrust which is equal to weight of to fluid displaced.

F = ρVg

V = F/ρg

where;

V is volume of water displacedρ is density of the immersed objectg is acceleration due to gravity

When an object with a significant weight is dropped into the pond, the level of water in the pond will rise and vice versa.

The son pops the helium balloon

The water level in the pond will be Unchanged, since the weight of the balloon is almost insignificant.

Fisherman knocks the tackle box overboard and it sinks to the bottom

The water level will rise because the weight of the tackle box is significant.

Son finds a cup and bails some water out of the bottom of the boat

The water level will fall since some volume of water is being removed.

Fisherman lowers himself in the water and floats on his back

The water level will be unchanged, since nothing new is added into the pond.

Fisherman lowers the anchor and it hangs one foot above the bottom of the pond

The water level will rise because the anchor will displace some volume of water upwards when it is dropped inside the pond.

Son gets in the water, looses his grip on the string, letting the balloon escape upwards.

The water level will rise, because the balloon will displace the water upwards when it escape upwards.

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A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​

Answers

A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

How to calculate  the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​?

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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If little herbert does something uncon
sciously he will probably

Answers

If little herbert does something unconsciously he will probably, he's definitely very unware of he has done

Doing something unconscious may have some implications which may not be favourable.

What is unconscious?

Unconscious simply means a state of someone being absent minded of what is happening to him or her at a particular period of time

So therefore, if little herbert does something unconsciously he will probably, he's definitely very unware of he has done

Complete question:

What happens if little herbert does something unconsciously he will probably

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An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a length of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)

Answers

The the tension in the horizontal rope is 7,019.4 N.

What is tension?

Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.

Tension can also be described as the action-reaction pair of forces acting at each end of said elements.

The unit of tension is Newton (same unit as force, since tension is also a force).

Tension in the horizontal rope

The tension in the horizontal rope is calculated as follows;

Apply the principle of torque;

T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ

T = (M + 2m)g sinθ/cosθ

T =  (M + 2m)g tanθ

let θ = 66⁰ (this value should be given in the question)

T =  (M + 2m)g tanθ

T = (74.9    +   2 x 122)(9.8)(tan 66)

T = 7,019.4 N

Thus, the the tension in the horizontal rope is 7,019.4 N.

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A 940-kg sports car collides into the rear end of a 2500-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
What was the speed sports car at impact?
Express your answer to two significant figures and include the appropriate units.

Answers

The speed of the sport car at the time of impact is 6.61 m/s.

What is the frictional force of the two cars?

Frictional force of the two cars = coefficient of kinetic frictin × mass × acceleration of gravity

= 0.8 × (2500+940) × 9.8

= 26970N

What is the acceleration of the skidded cars?As per Newton's second law of motion, force = mass × accelerationAcceleration= force / mass

= 26,970/3440

= 7.8 m/s²

What is the velocity of the sport car at the time of impact?As per Newton's equation of motion, V² - U² = 2aSHere, V = 0 m/s, a= -7.8 m/s², S= 2.8 mSo, 0²-U²= 2×(-7.8)×2.8

=> U = √43.68

= 6.61 m/s

Thus, we can conclude that the speed of the sport car at the time of impact is 6.61 m/s.

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