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Vaselesa [24]
3 years ago
10

What is the scientific term for this type of wave that was produced by a drum?

Physics
1 answer:
damaskus [11]3 years ago
7 0
Assuming this is the unit 8 waves project, it's a longitudinal wave, because the energy is going left to right.
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Where would you most likely find regional metamorphism? associated with a mountain belt associated with a volcano associated wit
vladimir1956 [14]
Regional metamorphism is a type of metamorphism in which the minerals and the textures of the rocks are changed or altered due to the effect of pressure and heat over a wide area or region.
This type of metamorphism often results in metamorphic rocks that are strongly foliated.
Metamorphism usually occurs in the folds of thrust mountain belts.

Answer: You would most likely find regional metamorphism associated with a mountain belt.

5 0
3 years ago
Read 2 more answers
A transverse wave on a string is described by the wave function
fgiga [73]

The acceleration of the given wave is 2.5m/s^2

<h3>Wave property</h3>

The standard wave function is expressed according to the equation

y = Asin(2πft+2πx/λ)

where

λ is the wave length

Given the equation below

y = 0.120 sin((π/8)x + 4πt)

Compare both equation to have:

(π/8)x = 2πx/λ

8 = 2/λ

λ = 2/8

λ = 1/4

λ = 0.25m

For the frequency

Compare both equation to have:

4πt = 2πft

4 = 2f

4/2 = 2f/2

f = 2 Hz

Speed = fλ

speed = 2(0.25)

speed = 0.5m/s

Determine the acceleration

acceleration = change in velocity/time

acceleration = 0.5/ 0.200

acceleration = 2.5m/s^2

Hence the acceleration of the given wave is2.5m/s^2

Learn more on wave function here: brainly.com/question/25699025

#SPJ4

7 0
2 years ago
7. Why are cooking pans made of aluminum? {Explain using the specific heat value of
Dimas [21]
Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.
6 0
2 years ago
A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea
grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

A = 1.803 ft

Period = \frac{\pi}{2} seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = \frac{w}{g}

m= \frac{32}{32}

= 1 slug

Also from Hooke's Law

2 k = 32

k = \frac{32}{2}

k = 16 lb/ft

Using the function:

\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0

x(0) = -1        (because of the initial position being above the equilibrium position)

x(0) = -6          ( as a result of upward velocity)

NOW, we have:

x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)

However;

x(0) = -1 means

-1 =c_1\\c_1 = -1

x(0) =-6 also implies that:

-6 =4(c_2)\\c_2 = - \frac{6}{4}

c_2 = -\frac{3}{2}

Hence, x(t) =-cos4t-\frac{3}{2} sin 4t

A = \sqrt{C_1^2+C_2^2}

A = \sqrt{(-1)^2+(\frac{3}{2})^2 }

A=\sqrt{\frac{13}{4} }

A= \frac{1}{2}\sqrt{13}

A = 1.803 ft

Period can be calculated as follows:

= \frac{2 \pi}{4}

= \frac{\pi}{2} seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

x* \frac{\pi}{2} = 4 \pi

x \pi = 8 \pi

x=8 cycles

5 0
3 years ago
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
3 years ago
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