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castortr0y [4]
2 years ago
12

Halle la velocidad de las ondas de una cuerda de violin de 820 g/m y 22cm de longitud, si la frecuencia de la fundamental es de

920 hz. Calcule la tensión de la cuerda
Physics
1 answer:
Mice21 [21]2 years ago
7 0

Answer:

your mom B on thezz NUTZ

Explanation:

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What determines if a material is a conductor or an insulator? Explain your answer and provide at least one example of each.
Dmitry [639]
A conducting material conducts or allows electricity to flow, while an insulator does not allow electricity to flow. For example think of a water pipe, if the pipe has a hole water can flow, on the other hand if it is just a solid rod, no water can flow through. I hope this helps.
4 0
3 years ago
A car drives 40 km due east and then 50 km due west .what is the cars overall displacement?
Firdavs [7]

Answer:10 km westExplanation:he go 40 east then 50 west 50-40 is 10 so he displaces 10 km and as west is more than east in terms of km so we will say that it's 10 km west pls mark as brainliest thanks

7 0
3 years ago
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What is the answer? Explain best and I will mark brainliest!
iVinArrow [24]
I would say B because it is near the ocean which can cause a tsunami but also because of the wind coming from the ocean (it might cause hurricanes and lots of storms) I’m not sure though but that’s what I think makes sense. Good Luck!
5 0
3 years ago
Read 2 more answers
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr
xxTIMURxx [149]
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
3 0
3 years ago
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