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3 years ago

What velocity will a freefalling object have after falling 80 meters

Physics

1 answer:

vovangra [49]3 years ago

6 0

Answer:

129.96

At a Gravitational acceleration of 32.17405 ( which is the normal rate for a freefall) you will geta velocity of 129.96 and the time of fall will be 4.039 seconds from 80 meters.

Why is the weight of a free falling body zero? It is not, an object in free fall will still have a weight, governed by the equation W = mg, where W is the object's weight, m is the object's mass, and g is acceleration due to gravity. Weight, however, has no effect on an objects free falling speed, two identically shaped objects weighing a different amount will hit the ground at the same time.

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Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE

Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s = 149597870700m/tiempo.

tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

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3 years ago

List two examples of situations were neutral object is charged by conduction,induction and friction

katovenus [111]

1. Boiling water
2. Hair straightener or curler

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3 years ago

(b) A cylinder of cross-sectional area 0.65m2 and

VLD [36.1K]

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19 $m^{3}$

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

$radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m$

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

$\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m$

Step Four:

calculate the volume of the cavity using volume = $\pi r^2 \times h$

Recall that the cavity has the same height as the original cylinder

$volume = \pi \times 0.115^2\times 0.32= 0.013m^3$

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3 years ago

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00

Masteriza [31]

The cart is moving by simple harmonic motion, and its position at time t is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the oscillation
$\omega$ is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released:
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
$\omega = \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s$

Therefore now we can calculate the position of the system at the time t=29.6 s:
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vovikov84 [41]

Answer:

what are you saying man. They do provide structure.

Explanation:

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4 years ago