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____ [38]
3 years ago
5

How does tightening a string on a guitar affect its natural frequency?

Physics
2 answers:
Flauer [41]3 years ago
6 0

Well, this is the answer in which, will be selected as a correct answer, on a answer choice selection: the natural frequency increases


iragen [17]3 years ago
3 0
By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.

Hope this helps
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an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul
kolbaska11 [484]

Answer:

r=\frac {mv}{qb}

Explanation:

In this case,  since the charged particle moves in circular motion,  the centripetal force is equivalent to the magnetic force.

\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}

5 0
3 years ago
A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met
MArishka [77]
The magnitude of the resultant is

         √ (22² + 2.2²)  =  √ (484 + 4.84)  =  √488.84  =  22.11 m/s .

The direction of the resultant is

         tan⁻¹(22N / 2.2E)  =  tan⁻¹(10)  =  5.71° east of north .
7 0
3 years ago
Which is correct? I need to know!!
Mnenie [13.5K]
What website are you using?
6 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0
3 years ago
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