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kotykmax [81]
3 years ago
13

A wave is produced in a rope. The wave has a speed of 33 m/s and a frequency of 22 Hz. What wavelength is produced? 0. 67 m 0. 7

5 m 1. 5 m 3. 0 m.
Physics
1 answer:
vekshin13 years ago
4 0

Answer:

Wavelength = <u>1.5 m</u>

Explanation:

The formula for waves in terms of wavelength, speed and frequency is:

Speed (v) = Frequency (f) × Wavelength (λ)

33 = 22 × λ

33 = 22λ

λ = \frac{33}{22}

So, λ = 1.5 m

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Please give me concept to solve this.
katrin [286]

Answer:

The difference in tension,  between adjacent sections of the pull cable at the given conditions is 17.701 kN

Explanation:

We take the cars as moving upwards such that the resultant pulling force on the car, F, along the cable is given by the relation

F_{car} = Upward tension force,  Tension_{(upwards)} - Downward tension force, Tension_{(downwards)} - Component of the weight of the car along the taut cable

The parameters given are;

Mass of car, m = 2750 kg

Angle of inclination of taut cables, θ = 35°

The upward acceleration of the car, a = 0.81 m/s²

Given that the weight is acting vertically downwards, we have;

Component of the weight of the car along the taut cable = m × g × sin(θ)

∴ Component of the weight of the car along the taut cable = 2750 × 9.81 × sin (35°) = 15473.66 N

We therefore have;

F_{car} =  Tension_{(upwards)} - Tension_{(downwards)} - 15473.66 N

F_{car}  = m × a = 2750 × 0.81 =  T_{upwards} - T_{downwards} - 15473.66

∴  Tension_{(upwards)} - Tension_{(downwards)} = 2750 × 0.81 + 15473.66 = 17701.16 N

Hence the difference in tension,  Tension_{(upwards)} - Tension_{(downwards)} between adjacent sections of the pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline = 17701.16 N or 17.701 kN.

5 0
3 years ago
A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is r
anygoal [31]

Answer:

Explanation:

General guidance

Concepts and reason

The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.

Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.

Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.

Fundamentals

The condition for diffraction maximum in the case of single slit diffraction is as follows:

sin Θ=λ/α

Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .

The incorrect options are as follows:

• The central bright fringe remains in same size.

The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.

• The effect cannot be determined unless the distance between the slit and screen is known.

Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.

• The central bright fringe becomes narrower.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.

If the width of the slit increases, then the central bright fringe width also increases.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.

The condition for diffraction maximum is as follows:

sin  Θ=λ/α

The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

Therefore,

• The central bright fringe becomes wider is correct.

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.  

Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.

If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.

Answer

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.

5 0
4 years ago
if the speed of a ball increased from 1m/s to 4 m/s, by how much would kinetic energy increase? .....help me please.
ivann1987 [24]
Can you write the formula for kinetic energy ?
Here, let me help you:

       Kinetic energy = (1/2) (mass) (speed)²  .

Look at the (speed) in the formula.  It's squared.

So if the speed gets multiplied by (something),
the kinetic energy gets multiplied by  (something)² .

If the speed starts out at 1 m/s, but gets multiplied by 4,
then the kinetic energy gets multiplied by  (4)²  =  16 .
7 0
3 years ago
What do we mean when we say that two light rays striking a screen are in phase with each other?
andrey2020 [161]

Answer:

When the electric field due to one is a maximum, the electric field due to the other is also a maximum, and this relation is maintained as time passes.

Explanation:

Phase of a wave or light ray is the instantaneous situation of the cycle in which the wave is at a given time.

When two waves are in phase means that the maximum and minimum of both coincide in time. They are in the same point of their cycle at the same time. And this relationship is maintained as time passes.

The waves can also be visualized as the oscillation of an electric field. (usually plotted like a sine function).

So the fact that two waves are in phase means that the maximums of their electric field coincide in time.

4 0
3 years ago
The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of
vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

=  4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0
4 years ago
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