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Zanzabum
4 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]4 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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How is an image produced by a plane mirror different than an image produced by a convex mirror?
katen-ka-za [31]

Answer:

D. It is upright.

Because convex mirror produce the real image, and the plane mirror produce it's upright.

7 0
4 years ago
Read 2 more answers
A grating has 380 lines/mm. How many orders of the visible wavelength 457 nm can it produce in addition to the m = 0 order? Plea
egoroff_w [7]

Answer:

5 orders.

Explanation:

The  condition for diffraction maxima,

sin \theta = mN\lambda\\m=\frac{sin \theta}{N\lambda}

m=\frac{sin 90^{\circ}}{(380\frac{lines}{mm}(\frac{1000mm}{1m} )) 457\times 10^{-9} m }\\m=5.75\simeq5

So maximum m is 5 which means it produce 5 orders.

Therefore, the number of possible order will produce by visible light of wavelength 457 nm is 5 orders.

6 0
4 years ago
a 100 kg person travels from sea level to an altitude of 5000 m. By how mans newtons does their weight change?
Bumek [7]

Answer:

Thus, the change in the weight of the person is 1.6N , option c is correct.

Explanation:

3 0
2 years ago
a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance
Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

v=u+at

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

t = 5 s is the time

So, the final velocity is

v=0+(2)(5)=10 m/s

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

v^2-u^2=2as

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

v = 10 m/s is the final velocity

Solving for s,

s=\frac{10^2-0^2}{2(2)}=25 m

3 0
3 years ago
A plane flies horizontally at an altitude of 3 km and passes directly over a tracking telescope on the ground. when the angle of
Lubov Fominskaja [6]
The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y 
Find x:
x = y/tan(θ) 
So x = 3/tan(π/6) 
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt 
Since altitude remains the same, dy/dt = 0. Now... 
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0 
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min 
3 0
4 years ago
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