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masha68 [24]
3 years ago
8

A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875

Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?
Physics
2 answers:
Fed [463]3 years ago
8 0

Answer:

a

  v _{max } =  6.82 \ m/s

b

 a_{max} =  37489.5 \  m/s^2

Explanation:

From the question we are told that

   The amplitude is A =  1.24 \  mm  =  1.24 * 10^{-3} \  m

   The frequency is  f =  875 \  Hz

   

Generally the maximum speed is mathematically represented as

           v _{max } =  A *  2 *  \pi * f

=>        v _{max } =  1.24*10^{-3} * 2 * 3.142 *  875

=>        v _{max } =  6.82 \ m/s

Generally the maximum acceleration is mathematically represented as

       a_{max} =  A  * (2 *  \pi *  f)

=>      a_{max} =  1.24*10^{-3} *  (2 * 3.142 *  875 )^2

=>       a_{max} =  37489.5 \  m/s^2

mario62 [17]3 years ago
5 0

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

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frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0
3 years ago
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If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated
sp2606 [1]

Answer:

611.8 nm

Explanation:

n_{oil} = Index of refraction of soap bubble  = 1.33

t = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

\lambda = wavelength of light = ?

m = order = 0

For reflection , the necessary condition is

2 n_{oil} t = (m + 0.5) \lambda

2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda

\lambda = 6.118\times 10^{-7}

\lambda = 611.8 nm

7 0
2 years ago
If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you
Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is t_1 and when he swims in downward direction is t_2

Lets distance is d and it will be remain constant in both the case

d=(u+v)t_1

d=(u-v)t_2

(1.73+0.52)t_1=(1.73-0.52)t_2

t_2=1.85t_1

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

t_1=0.44d\ sec

t_2=0.82d\ sec

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

percentage\ time =\dfrac{1.28-1.15}{1.15}

 Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0
3 years ago
A 10000 N net force is accelerating a car at a rate if 5.5 m/s^2. What is the cars’s mass?
lianna [129]

Answer:

1800 kg

Explanation:

Newton's second law:

F = ma

10000 N = m (5.5 m/s²)

m ≈ 1800 kg

8 0
3 years ago
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Four resistors , R1=12ohms, R2= 15 ohms , R3= 18 ohms and R4=10 ohms are connected to 6 v battery in series what is the total re
morpeh [17]

Answer:

55 ohms

Explanation:

if it was series circuit, then you just need to add all the resistances

6 0
3 years ago
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