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svlad2 [7]
3 years ago
12

A taut rope has a mass of 0.123 kg and a length of 3.54 m .What average power must be supplied to the rope to generate sinusoida

l waves that have amplitude 0.200 mm and wavelength 0.600 mm if the waves are to travel at 22.0 m/s ?
Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

811.54 W

Explanation:

Solution

Begin with the equation of the time-averaged power of a sinusoidal wave on a string:

P = \frac{1}{2\\} μ.T².ω².v

The amplitude is given, so we need to calculate the linear mass density of the rope, the angular frequency of the wave on the rope, and the frequency of the wave on the string.

We need to calculate the linear density to find the wave speed:

μ = \frac{M}{L} = 0.123Kg/3.54m

The wave speed can be found using the linear mass density and the tension of the string:

v= 22.0 ms⁻¹

v = f/λ =  22.0/6.0×10⁻⁴

= 36666.67 s⁻¹

The angular frequency can be found from the frequency:

ω=  2πf=2π(36666.67s−1)  = 2.30 ×10⁻⁵s⁻¹

Calculate the time-averaged power:

P =\frac{1}{2}μΤ²×ω²×ν

= \frac{1}{2} ×( 0.03475kg/m)×(0.0002)²×(2.30×10⁵)² × 22.0

= 811.54 W

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Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th
Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0
3 years ago
I need help with the circled one
Whitepunk [10]
Yes the answer is yes 
7 0
3 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

\rho(r) = ar^2 - br^3

r is the radius of the spherical shell

dr is the thickness

volume of shell

dV = 4 \pi r^2 dr

mass of shell

dM = \rho(r)dV

\rho = \rho_0 - br

now,

dM = (\rho_0 - br)(4 \pi r^2)dr

integrating both side

M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr

M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

7 0
3 years ago
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d
Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0
3 years ago
Whats the answer???????????????????????
murzikaleks [220]

The answer is 30 ... same as the Atomic number.

4 0
3 years ago
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