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Travka
4 days ago
12

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

function of time according to the equation y = 0.049 cos(7t) where y is the displacement in meters and the time t is in seconds. show answer No Attempt 33% Part (a) What is the amplitude of the wave, in meters?
Physics
1 answer:
kicyunya [1K]4 days ago
3 0

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that

A harmonic wave propagates in the positive x direction at 6 m/s along a tight string. A fixed point along this string oscillates over time according to the equation:

y = 0.049 \cos(7t).......(1)

The general wave equation is expressed as:

y=A\cos(\omega t).......(2)

A denotes the wave's amplitude

When we compare equation (1) with (2), we find:

A = 0.049 meters

Thus, the amplitude of the wave is 0.049 meters.

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The wavelength of light is 5000 angstrom. Express it in nm and m.
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Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

1 \:  angstrom = 1 \times  {10}^{ - 10} m

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

7 0
13 days ago
A 68 kg hiker walks at 5.0 km/h up a 9% slope. The indicated incline is the ratio of the vertical distance and the horizontal di
Softa [913]

The formula used to calculate power is:

Power = Work / Time

Where Work is defined as Force multiplied by Distance, hence:

Power = Force * Distance / Time

This can also be expressed as:

Power = Force * Velocity<span>


Converting the velocity from km/h to m/s:</span>

Velocity = (5 km / h) (1000 m / km) (1 h / 3600 s)<span>
Velocity = 1.39 m/s 

</span>

The calculated force corresponds to the vertical component of the hiker's weight:

Force = Wy * g = W * sin θ * g

 

To find the angle θ, we utilize the slope, which is defined as:

slope = tan θ = ratio of vertical to horizontal distance

tan θ = 0.09

θ = 5.14˚

Thus, we have Force = 68 kg * sin(5.14˚) * 9.8 m/s²<span>
Force = 666.4 * sin(5.14) = 59.73 N 

</span>

<span>Now, let's compute the power:
<span>Power = 59.73 N * 1.39 m/s = 82.96 W
Given the hiker's efficiency is at 25%, we can find the metabolic power: 
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6 0
2 days ago
In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the
Sav [1105]
The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.

If this is not correct, feel free to inform me and I will find out the right answer. However, I am confident this is accurate.:) </span>
8 0
2 days ago
Read 2 more answers
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
serg [1198]

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

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7 0
11 days ago
A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Sav [1105]

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

7 0
3 days ago
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