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Zanzabum
3 years ago
12

B. A guitar string has a length of 0.90 m. When you pluck it, it plays a “C” that has a frequency of 256 Hz. How fast is the wav

e moving back and forth along the string? (Choose one of the equations from the box at the top of the page to solve this problem.)
Equations to use: v= λ ∙ f v=d/t
Physics
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

461  m/s

Explanation:

Given the string length of guitar, l = 0.90 m

Frequency , f = 256 Hz

Using frequency, wavelength and velocity relation as

v=f λ

Here λ is wavelength.

Now for guitar string the length must be equal to half of wavelength, so

λ = 2l

Therefor above formula becomes as  

v = f×2l

Substituting the given values, we get

v =256×2×0.90

v = 460.8 m/s = 461 m/s

Thus, the wave moving back and forth along the string with velocity 461 m/s.

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In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
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A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

L = inductance

Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

Hence, The time is 0.5 sec.

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Fofino [41]
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Pressure=P
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(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.
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A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca
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i believe it is C....tell me if im right please<3

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