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2 years ago

Which term best describes the motion of the rope particles in relation to the motion of the rope wave shown in the photograph

Physics

2 answers:

Nady [450]2 years ago

3 0

Answer:

A: Perpendicular

Explanation:

The question is incomplete as it lacks the image of the rope wave motion.

However, as found on "estudyassistant", the options are;

A) Perpendicular

B) Circular

C) Longitudinal

D) Parallel

From all that, we can say that;

The rope's are moving simultaneously in the same pattern without touching each other.

This is therefore a mechanical wave being created with the motion having oscillations that are perpendicular to the direction of energy transfer of the ropes.

This is a definition of transverse waves because the rope particle motion is perpendicular to the wave motion.

Ivenika [448]2 years ago

3 0

Answer:

A: Perpendicular

Explanation:

Read above explanation.

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A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

denis-greek [22]

The radius of the cylinder is equal to half the diameter:

$r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm$

The volume of the cylinder is given by:

$V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3$

where h is the heigth of the cylinder. Converting into meters,

$V=1.28 \cdot 10^{-4} m^3$

And the density of the material will be given by the ratio between the mass and the volume:

$d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3$

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3 years ago

Coronary artery disease is a disease of the heart where the arteries and blood vessels become clogged with fat deposits called _

sineoko [7]

Plaque is the reasoning for this question ;)

8 0

3 years ago

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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr

nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

$\frac{E}{t} = VI$

solving for t

$t = \frac{E}{VI}$

$t = \frac{m_w C_w \Delta T}{VI}$

$t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}$

t = 444.125 sec

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3 years ago

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All of the waves in the electromagnetic spectrum are _______ waves.

user100 [1]

Transverse, I think. I may be wrong.

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3 years ago

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A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at

Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)$

Rearranging terms, we have:

$m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)$

We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

$\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

Rearranging , and simplifying common terms, we have:

$m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

$v_{10} + v_{1f} = v_{2f}$

Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

$m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$

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3 years ago