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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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A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset vo
kiruha [24]

Answer:

The value is  V_{os} = 0.001 \  V

Explanation:

From the question we are told that

     The circuit resistance is  R_1 =  10 \ k \Omega

     The feedback resistance  is  R_f =  220 \ k \Omega

      The offset current is  I_{os } = 100 \  nA  =  100 * 1)^{-9} \ A

Generally the offset voltage is mathematically reparented as

           V_{os} =  R_f * I_{os}

=>        V_{os} = 10 *10^{3}*  100 *10^{-9}

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6 0
3 years ago
A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
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g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e
rusak2 [61]

Answer:

(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0

Given the function:

p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in [0,\infty).

(2)

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The function p(x) satisfies the conditions for a probability density function.

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