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Igoryamba
3 years ago
9

Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 9.70 m apart. If they shake the

bridge twice per second, what is the propagation speed of the waves (in m/s)?
Physics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

<em> The speed of the wave is 19.4 m/s</em>

Explanation:

The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m

The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.

this means that the frequency f = 2 Hz

The speed of a wave = fλ = 9.70 x 2 =<em> 19.4 m/s</em>

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What is the weight of a spring balance when the point is 30​
andrezito [222]

Answer:

A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)

Explanation:

8 0
2 years ago
As 390 g of hot milk cools in a mug it transfers 30 000 j of heat to the environment. whats is the temperature change of the mil
Fed [463]
You have to use the specific heat equation. 

Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.

So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

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I'm assuming the temperature is C, since it was not specified.

Hope this helps!
8 0
3 years ago
Why is it not necessary for radio telescope surfaces to be as smooth as a mirror?
Law Incorporation [45]
It doesn't on account of radio waves are longer than optical waves. Radio waves are a sort of electromagnetic radiation with wavelengths in the electromagnetic range longer than infrared light. These long waves are in the radio locale of the electromagnetic range.
8 0
3 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0
3 years ago
An RLC circuit is used in a radio to tune into the radio lagos fm Station broadcasting at 93.5Hz. The resistance is 15ohms and t
bazaltina [42]

The characteristics of the RLC circuit allow to find the result for the capacitance at a resonance of 93.5 Hz is:

  • Capacitance is C = 1.8 10⁻⁶ F

A series RLC circuit reaches the maximum signal for a specific frequency, called the resonance frequency, this value depends on the impedance of the circuit.

            Z^2 = R^2 + ( wL - \frac{1}{wC} )^2  

Where Z is the impedance of the circuit, R the resistance, L the inductance, C the capacitance and w the angular velocity. The negative sign is due to the fact that the current in the capacitor and the inductor are out of phase.

In the case of resonance, the impedance term completes the circuit as a resistive system.

           wL - \frac{1}{wC} = 0  \\w^2 = \frac{1}{LC}  

           

Indicate that the inductance L = 1.6 H and the frequency f = 93.5 Hz.

Angular velocity and frequency are related.

         

         w = 2π f

           

Let's  substitute.

          C = \frac{1}{L ( 2 \pi f)^2 }  

 

Let's calculate.

         C = \frac{1}{1.6 \ ( 2\pi \ 93.5)^2}  

         C = 1.8 10⁻⁶ F

In conclusion with the characteristics of the RLC circuits we can find the result for the capacitance at a 93.5 Hz resonance is:

  • Capacitance is C = 1.8 10⁻⁶ F

Learn more about serial RLC circuits here: brainly.com/question/15595203

4 0
2 years ago
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