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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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3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

Irina-Kira [14]

Answer:

35.6 m

Explanation:

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3 years ago

A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat

Veronika [31]

Answer:4.9m/s^2

Basically answer A

Explanation:

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3 years ago

An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b

Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

8 0

3 years ago

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

= $\frac{(-392 + 2940)}{12}$ = 212.33 N

5 0

3 years ago