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Fofino [41]
3 years ago
12

You are making a telephone out of two aluminum cans and some string. You can choose between two types of string: a 2-m length of

thin cooking twine or a 2-m length of heavy rope, both used at the same tension. Which type of string should you use so that you have the fastest wave speed in your telephone?
Physics
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

C)You should use the thin cooking twine.

Explanation:

A)You can choose either because they are the same length and will produce the same wave speed.

B)You should use the heavy rope.

C)You should use the thin cooking twine.

The speed of wave in a string is given by the following formula:

|v| = \sqrt{\frac{F_T}{u} }

Where |v| = speed of wave, F_T = tension in the string, and μ = mass per length of the string.

<em>Even though the two strings have the same length, the μ (mass/length) for the heavy rope will be more than the that of a thin rope. Consequently, the </em>F_T<em>:μ for the thin rope will be higher than that of the heavy rope and as such, gives a bigger |</em>v<em>|. </em>

Therefore, the thin rope should be used in order to get a faster wave speed in the telephone.

The correct option is C.

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20 N north because He is pushing it 20 newtons north.

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2 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

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C. 110 - 140

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The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at the bottom

K_f is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

mgh_i + K_i = mgh_f + K_f

where:

m = 0.5 kg is the mass of the apple

g=9.8 m/s^2 is the acceleration of gravity

h_i = 2 m is the initial height

h_f=1.50 m is the final height

The initial kinetic energy is zero, since the ball starts from rest:

K_i = 0

Therefore we can solve the equation for K_f, the final kinetic energy of the ball:

K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

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