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olga nikolaevna [1]
3 years ago
14

Two Velocities in a Traveling Wave? Wave motion is characterized by two velocities: the velocity with which the wave moves in th

e medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).
Consider a transverse wave traveling in a string. The mathematical form of the wave is
y(x,t) = A sin(kx-omega...
Physics
1 answer:
Lyrx [107]3 years ago
3 0

For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0. 
</span>
 y(x,t) = A sin(kx-omega t) 
d{y(x,t)}/d{x} = A k cos(kx - omega t)
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Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,
makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/s^{2}.

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e V^{2} = U^{2} ± 2as

where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.

From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.

So that since Poala flies against gravity, then we have:

V^{2} = U^{2} - 2as

0 = (4)^{2} - 2(a x 0.201)

  = 16 - 0.402a

0.402a = 16

a = \frac{16}{0.402}

  = 39.801

a = 39.80 m/s^{2}

Therefore Paola's acceleration is 39.80 m/s^{2}.

Visit: brainly.com/question/17493533

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True or False:
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This is a true statement
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Para que sirve la caida libre
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3 years ago
How much energy is given to each coulomb of charge passing through a 6 V battery
ELEN [110]
1 volt = 1 joule per coulomb. Current doesn't actually pass 'through' a battery. But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
4 0
3 years ago
A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

4 0
3 years ago
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