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yan [13]
3 years ago
8

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 0.120 m. The mass is now displ

aced from its equilibrium position and undergoes simple harmonic oscillations. How long does it take the mass to complete one full oscillation?
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Time period of the oscillation will be 0.695 sec

Explanation:

We have given mass attached to the string m = 0.350 kg

The spring is stretched by 0.120 m

So x = 0.120 m

We know that force is given by F = Kx

So mg = kx

0.350\times 9.8=k\times 0.120

k = 28.5833

Now time period is given by

T=2\pi \sqrt{\frac{m}{k}}=2\times 3.14\sqrt{\frac{0.35}{28.5833}}=0.695sec

So time period of the oscillation will be 0.695 sec

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A disk is rotating with an angular velocity function given by ω=Kt+L. What is the angular acceleration of the disk at t=T ?
PolarNik [594]

Answer:

Angular acceleration of the disk ∝ = K

Explanation:

Given that;

Angular velocity function given by ω = Kt + L

Angular acceleration of the disk at t=T is = ?

Now,

Angular velocity;

ω = Kt + L

Angular acceleration is;

∝ = dω / dt

= \frac{d}{dt} ( Kt + L )

At t = T

∝ = K

Because  ∝  is not dependent on t

Therefore

Angular acceleration of the disk ∝ = K

8 0
3 years ago
Pluto orbits the Sun at an average distance of 5.91 × 10^12 meters. Pluto’s diameter is 2.30 × 10^6
irakobra [83]

Well first of all, since the question deals with the relationship of Charon and Pluto, we don't care about their distance from the sun, because that has no effect on the gravitational forces between them.

All we need is Newton's law of universal gravitation:

                             Gravitational forces = G  m₁ m₂ / D²    .

' G ' is the gravitational constant ...  6.67 x 10⁻¹¹ newt-m²/kg²

m₁ = either one of the masses ...  1.31 x 10²² kg  (Pluto)

m₂ = the other mass ... 1.55 x 10²¹ kg  (Charon)

D = the distance between their centers ... 1.96 x 10⁷ meters

   Gravitational forces = G  m₁ m₂ / D²

     =  (6.67 x 10⁻¹¹ newt-m²/kg²)
         times (1.31 x 10²² kg) times (1.55 x 10²¹ kg)
         divided by  (1.96 x 10⁷ meters)²

      =  (6.67 x 1.31 x 1.55 x 10⁻¹¹ ⁺ ²² ⁺ ²¹ ⁻ ¹⁴) / (1.96)²

      =  (13.54 x 10¹⁸) / (3.84)  =  3.53 x 10¹⁸ newtons .

<span>This is the gravitational force of attraction that Pluto exerts
on Charon.  It's also the </span><span>gravitational force of attraction that
Charon exerts on Pluto.  The gravitational forces are always
equal and opposite.
 </span>         




4 0
3 years ago
El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es
Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión (P), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

P = \frac{F}{A}

Donde:

F - Fuerza motriz, medida en newtons.

A - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón (A), medido en metros cuadrados, es determinado por:

A=\frac{\pi}{4}\cdot D^{2}

Donde D es el diámetro del pistón, medido en metros.

Si D = 0.32\,m y F =17,640\,N, entonces la presión neumática es:

A = \frac{\pi}{4}\cdot (0.32\,m)^{2}

A \approx 0.080\,m^{2}

P = \frac{17,640\,N}{0.080\,m^{2}}

P = 220,500\,Pa

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0
3 years ago
Tracy scuffs her socked feet across a carpet. When she touches a doorknob, she gets a small shock. How did Tracy become charged?
densk [106]
<span>through friction between her feet and the carpet </span>
5 0
4 years ago
Read 2 more answers
Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t
ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body       Mass (gram)     Engine      Force (newtons)

1                   500                 1                     25

2                  1500                2                    20

3                  750                  3                    30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

 20    = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

                      $=\frac{200}{5}$

                      $= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $ 40 \ m/s^2$ at the start of the launch.

8 0
3 years ago
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