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Lubov Fominskaja [6]
3 years ago
10

A violin has a string of length

Physics
2 answers:
Shtirlitz [24]3 years ago
7 0

Answer:

1.9kHz

Explanation:

Given data

wavelength \lambda= 0.32m

velocity v= 622 m/s

We know that

v= f* \lambda\\\\f= v/ \lambda

substitute

f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz

Hence the frequency is 1.9kHz

Kazeer [188]3 years ago
7 0

Answer:

971.2

Explanation:

It was right on acellus :)

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How long does it take an object to travel 375 m at a rate of 25 m/s?
Leni [432]

Answer:

15s

Explanation:

Given parameters:

Distance traveled  = 375m

Speed = 25m/s

Unknown:

Time taken = ?

Solution:

To solve this problem, we make time the subject of the speed equation.

    Speed  = \frac{distance}{time}  

  Time  = \frac{distance}{speed}  

 Now insert the parameters and solve;

  Time  = \frac{375}{25}   = 15s

3 0
3 years ago
Calculate the wavelength of light that has a frequency of 5.2 x 1012 1/s.
Travka [436]

Answer:

Wavelength = 5.77 * 10^-5 meters.

Explanation:

Given the following data:

Frequency of light = 5.2 *10^12 Hz

We know that the Speed of light = 3.0 * 10^8 m/s

To find the wavelength of light;

Mathematically, wavelength is calculated using this formula;

Wavelength = \frac {speed}{frequency}

Substituting into the equation, we have;

Wavelength = \frac {3*10^{8}}{5.2 *10^{12}}

Wavelength = 5.77 * 10^-5 meters.

6 0
3 years ago
Producers,____________, and_______________ help to move matter and energy through ecosystems.
Phoenix [80]
Producers, consumers, and decomposers help to move matter and energy through ecosystems.

Hope this helps! :)
3 0
3 years ago
When a cannon is fired, how does the size of the force of the cannon on the cannonball compared with the force of the cannonball
kati45 [8]
The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
3 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
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