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3 years ago

Cual es l diferencia entre ruido y sonido

Physics

1 answer:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

E.l soni.do es un.a sensac.ión, en el órg.ano del oído, prod.ucida por el movimie/nto ondu>latorio de un m/edio elástico (normal/mente el aire), debi.do a ra.pidísimos ca/mbios de pre.sión, generado/s por el movimiento vibrat.orio d.e un cuerpo sonoro. ... /El ruido se consid/era a to/do sonid.o / o no de.seado.

Explanation:

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A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it

Mashutka [201]

Answer:

9 joules of heat energy was produced

Explanation: there is no acceleration therefore its not a kinetic energy

Energy= force × distance

= 3×3

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3 years ago

An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1

Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0

3 years ago

Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i

UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

= 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

V = k q/r

Where,

= 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

V₁ = 9 x 10⁹ x 30 / 0.3

= 1.5 x 10¹² J

And,

V₂ = 1.5 x 10¹² J

The energy stored in it is,

W = V₂ - V₁

= 0

Hence, the energy stored in the charge is, W = 0

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3 years ago

A student made a model of isostasy by placing a block of wood in a beaker of water. What does the water in the beaker represent

Setler [38]

The answer is C. the earths mantle because the wood in this case is the surface

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3 years ago

Read 2 more answers

A factory worker pushes a 32.0 kgkg crate a distance of 4.0 mm along a level floor at constant velocity by pushing horizontally

Luda [366]

Answer:

a) The force that the worker must apply has a magnitude of 75.317 newtons.

b) The external force does a work of 0.301 joules.

c) The friction force does a work of -0.301 joules.

d) Both normal force and gravity have done a work of 0 joules.

e) The total work done on the crate is 0 joules.

Explanation:

a) As the crate is moving at constant velocity, we know that magnitude of the force done on the crate must be equal to the friction force. Hence, we must use the following formula:

$F = \mu\cdot m\cdot g$ (1)

Where:

$F$ - External force, in newtons.

$\mu$ - Coefficient of kinetic friction, no unit.

$m$ - Mass, in kilograms.

$g$ - Gravity acceleration, in meters per square second.

If we know that $\mu = 0.24$ , $m= 32\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the external force is:

$F = (0.24)\cdot (32\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 75.317\,N$

The force that the worker must apply has a magnitude of 75.317 newtons.

b) The direction of the force is parallel to the direction of motion. The work done by this force ( $W_{F}$ ), in joules, is determined by this formula:

$W_{F} = F\cdot \Delta s$ (2)

Where $\Delta s$ is the distance travelled by the crate, in meters.

If we know that $F = 75.317\,N$ and $\Delta s = 0.004\,m$ , then the work done by the force is:

$W_{F} = (75.317\,N)\cdot (0.004\,m)$

$W_{F} = 0.301\,J$

The external force does a work of 0.301 joules.

c) The direction of the friction force is antiparallel to the direction of motion. The work done by this force ( $W_{f}$ ), in joules, is determined by this formula: