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dybincka [34]
3 years ago
6

Three diffrent examples of accelerated motion

Physics
1 answer:
LekaFEV [45]3 years ago
3 0

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

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If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
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Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

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3 years ago
A wave has a wavelength of 3m and a frequency of 10Hz. How fast is the wave traveling
andreev551 [17]
Firstly, do you understand the concept of a wavelength?
7 0
3 years ago
For our statistical definition of entropy, we use the equation S = klnW. In this equation, what does k represent?
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<span>In the </span>natural logarithm<span> format or in equivalent notation (see: </span>logarithm) as:


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I hope this helps. </span>
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3 years ago
Ozone gas absorbs the ultraviolet radiations of the sun in the
bija089 [108]
Hello,

The answer is option C "<span>thermosphere".</span><span>

Reason:

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7 0
3 years ago
Read 2 more answers
Una persona cierra una puerta de 1 metro de ancho aplicando una fuerza de 40 [N], perpendicular a ella, a 90 [cm] de su eje de r
Damm [24]

Answer:

El módulo del torque aplicado es 36 Nm

Explanation:

En los movimientos rotatorios, la cantidad de fuerza aplicada no depende de la acción gravitacional sino del momento inercial, que es el equivalente angular de la inercia (masa) y representa la resistencia que un objeto ofrece al rotar alrededor de su eje. Cuando un cuerpo rígido rota alrededor de su eje debe considerarse , además de la masa, el radio de giro ya que estos dos factores determinan la resistencia del cuerpo a los cambios de movimiento rotatorio a través de un eje determinado.

De esta manera, se llama torque o momento de una fuerza a la capacidad de dicha fuerza para producir un giro o rotación alrededor de un punto.

En muchas ocasiones el punto de aplicación de la fuerza no coincide con el punto de aplicación en el cuerpo. En este caso la fuerza actúa sobre el objeto y su estructura a cierta distancia, mediante un  elemento que traslada esa acción de esta fuerza hasta el objeto. Entonces, el momento de una fuerza  es, matemáticamente,  igual al producto de la intensidad de la fuerza (módulo) por la distancia desde el punto de aplicación de la fuerza hasta el eje de giro:

M=F*d*sen θ

donde F es la fuerza en Newton (N), d la distancia en metros (m), θ el ángulo que forma la fuerza con el objeto al cual se le aplica la fuerza y M el momento, que se mide en Newton por metro (Nm).

En este caso:

  • F= 40 N
  • d= 90 cm= 0.9 m (siendo 100 cm= 1 m)
  • θ= 90° ya que la fuerza se aplica de forma perpendicular. Entonces sen θ= sen 90= 1

Reemplazando:

M=40 N*0.9 m* 1

Resolviendo:

M= 36 Nm

<u><em>El módulo del torque aplicado es 36 Nm</em></u>

3 0
3 years ago
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