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3 years ago

Which statements accurately describe mechanical waves? Check all that apply.

Physics

2 answers:

Lorico [155]3 years ago

4 0

Answer:

-Energy is transferred through vibrating particles

-An ocean wave moving through water is an example of a mechanical wave

-A longitudinal wave is a type of mechanical wave.

-A transverse wave is a type of mechanical wave

Explanation:

stiks02 [169]3 years ago

3 0

Answers that apply include

Energy is transferred through vibrating particles
An ocean wave moving through water is an example of a mechanical wave
A longitudinal wave is a type of mechanical wave.
A transverse wave is a type of mechanical wave

Mechanical waves can't pass through vacuum like electromagnetic waves because t depends on the transfer of energy between particles of matter (matter that has inertia and elasticity). This energy propagates in the same direction as the wave. Another example of mechanical energy is sound. In addition to longitudinal and transverse waves, another type of mechanical wave is surface waves.

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What best describes an impulse acting on an object?

Triss [41]

The force on an object use Socratic glad to help ..

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3 years ago

The literary movement with which Wilde is most closely associated is Select one: a. romanticism b. existentialism c. Art for art

jeyben [28]

Answer:

Correct answer is ''c'' Art for art's Sake

Explanation:

Wilde wrote in the literary movement called Aestheticism during the late nineteenth century. Contrary to popular belief, Wilde did not create the literary movement, but played a role as a leader who promoted the movement. While Wilde was a college student the works of Algernon Charles Swinburne and Edgar Allan Poe influenced his own writing style. Also, the English essayist Walter Pater helped to form Wilde's humanistic aesthetics.

The philosophical foundations of Aestheticism come from Immanuel Kant. He formulated the idea of "art for art's sake". He believed that art was to be enjoyed for its own beauty regardless of social or moral concerns.

6 0

3 years ago

contractor will use a package emulsion having a specific gravity of 1.25 and relative bulk strength of 130 to open an excavation

tester [92]

Answer:

The burden distance is 7 ft

Solution:

As per the question:

Specific gravity of package emulsion, $SG_{E} = 1.25$

Specific gravity of diabase rock, $SG_{R} = 2.76$

Diameter of the packaged sticks, d = 3 in

Now,

To calculate the first trail shot burden distance, B:

$B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d$

$B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22$

B = 7 ft

5 0

3 years ago

A cylinder that is 15 cm tall is filled with water. If a hole is made in the side of the cylinder, 4.0 cm below the top level. A

olga55 [171]

Answer:

About 13 cm

Explanation:

We are given that

Length of cylinder,l=15 cm= $15\times 10^{-2} m$

1 m=100 cm

h=4 cm= $\frac{4}{100}=0.04 m$

Horizontal velocity of the existing water stream= $v=\sqrt{2gh}$

Where $g=9.8m/s^2$

Using the formula

$v=\sqrt{2\times 9.8\times 0.04}=0.885m/s$

h'=l-h=15-4=11 cm= $\frac{11}{100}=0.11 m$

Initial velocity of water stream,u=0

$t=\sqrt{\frac{2h'}{g}}=\sqrt{\frac{2\times 0.11}{9.8}}=0.15 s$

Distance, d=vt

Using the formula

$d=0.885\times 0.15=0.133 m$

$d=0.133\times 100=13.3 cm\approx 13 cm$

6 0

3 years ago

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

$\displaystyle K=\frac{1}{2}m.v^2$

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$

$\displaystyle h_M=6\frac{v^2}{2g}=6h$

This means the object could fall from six times the original height and still be safe

6 0

3 years ago