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FinnZ [79.3K]
2 years ago
9

A sinusoidal wave in a string is described by the wave functiony=0.150 sin (0.800x - 50.0t)where x and y are in meters and t is

in seconds. The mass per length of the string is 12.0 g/m(a) Find the maximum transverse acceleration of an element of this string
Physics
1 answer:
AleksAgata [21]2 years ago
6 0

The maximum  transverse acceleration of an element of this string is 375 m/s

<h3>What is the  maximum transverse acceleration?</h3>

For a transverse wave, the velocity is known to be perpendicular to the direction of the propagation of the particular wave. To get the acceleration, what is necessary is to take the partial  derivative as it compares to time and the velocity.

With this in mind, we would have

transverse acceleration

a(t) = d²y/dt² = -0.150*50.02sin(0.8x - 50t)

= 0.150*502 = 375 m/s2

Hence we can say that the maximum acceleration of the element on the string is given as 375 m/s²

Read more on acceleration here

brainly.com/question/605631

#SPJ4

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Marcus attends a night program at an observatory to learn more about the solar system. He learns that the solar system contains
Lesechka [4]

Answer: D meteoroid

Explanation:

Meteoroid is the smallest among them. It is a tiny asteroid or the broken-off crumb of comets and sometimes planets. It ranges in size from a grain of sand to boulders 3 feet (1 meter) wide. When meteoroids collide with a planet's atmosphere, they become meteors. If those meteors survive the atmosphere and hit the planet's surface, their remains are called meteorites

8 0
3 years ago
An FM radio station broadcasts at 9.23 × 107 Hz. Given that the radio waves travel at 3.00 × 108 m/s, what is the wavelength of
STALIN [3.7K]

Answer:

3.2m

Explanation:

Given parameters:

Frequency of the FM radio = 9.23 x 10⁷Hz

Velocity of the waves = 3  x 10⁸m/s

Unknown:

Wavelength of the wave = ?

Solution:

To solve for the wavelength of the wave, we need the velocity equation;

       Velocity = frequency x wavelength.

Radio waves are all electromagnetic radiations produced by both electrical and magnetic fields perpendicularly oriented to one another.

 Since the unknown is wavelength, we solve for it:

       

    3  x 10⁸  = 9.23 x 10⁷ x wavelength

          wavelength = \frac{ 3 x 10^{8} }{9.23 x  10^{7} }

        wavelength = 3.2m

7 0
4 years ago
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

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5 0
3 years ago
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

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p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

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Equating initial and the final momenta we get

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Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

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v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

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How does pressure affect surface tension
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the effect of pressure on surface tension can be attributed in part to absorption of gas at the surface of the liquid and in part to an intrinsic decrease in density of the liquid in the neighborhood of the surface.

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3 0
3 years ago
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