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Lorico [155]
3 years ago
5

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

ched. The mass is released and falls, stretching the spring by 20 cm before coming to rest at its lowest point. It then continues to oscillate vertically.a. What is the spring constant? (K=)
b. What is the amplitude of the oscillation?
c. What is the frequency of the oscillation?
Physics
1 answer:
Nataly [62]3 years ago
5 0

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π \sqrt{\frac{k}{m} }

f= 1/2π \sqrt{\frac{32.34}{0.33} }

f= 1.57

f≈ 1.6 Hz

Therefore,  the frequency of the oscillation is 1.6 Hz

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3 years ago
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Se colocan tres objetos, muy cerca uno del otro, dos al mismo tiempo. Cuando se juntan los objetos A y B, se repelen. Cuando se
docker41 [41]

Answer:

Los objetos A y C tienen cargas del mismo signo (opcion a)

Explanation:

Hay dos tipos de cargas : cargas positivas y cargas negativas.

La Ley de Coulomb dice que la fuerza electrostática entre dos cargas puntuales es proporcional al producto de las cargas e inversamente proporcional al cuadrado de la distancia que las separa, y tiene la dirección de la línea que las une y se cumple que:

  • La fuerza ejercida sobre una carga apunta hacia la otra cuando las dos tienen distinto signo (fuerza atractiva).
  • El sentido de la fuerza se dirige hacia el lado opuesto de la carga cuando ambas tienen el mismo signo (fuerza repulsiva).

Es decir que las cargas de igual signo se repelen, mientras que las de diferente signo se atraen.

Entonces, si se juntan los objetos A y B y se repelen significa que la carga es del mismo signo.

Cuando se acercan los objetos B y C, se repelen. Entonces significa que posee carga de igual signo.

Por lo que podes concluir que <u><em>los objetos A y C tienen cargas del mismo signo (opcion a)</em></u>

8 0
3 years ago
Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum
marta [7]

Answer:

  • v_{top} = 61.96 \frac{mi}{h}

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

v_{top} = 27.7 \frac{m}{s}

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

1609.34 \ m = 1 \ mi

Now, we can divide by 1609.34 meters on both sides:

\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}

The left sides equals 1, so

1 = \frac{1 \ mi}{ 1609.34 \ m}

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 0.0172120 \frac{mi}{s}

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

1 \ h = 3600 \ s

as the seconds are dividing in the velocity, we know divide by 1 hour.

\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}

1 = \frac{3600 \ s}{ 1 \ h}

and know we just multiply our top speed by this conversion factor

v_{top} = 0.0172120 \frac{mi}{s}  \frac{3600 \ s}{ 1 \ h}

v_{top} = 61.96 \frac{mi}{h}

8 0
3 years ago
By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l
Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient \alpha=23.0\times10^{-6}/C^{\circ}

We need to calculate the volume expansion coefficient

Using formula of  volume expansion coefficient

\beta=3\alpha

Put the value into the formula

\beta=3\times23.0\times10^{-6}

\beta=0.000069=69\times10^{-6}/C^{\circ}

We need to calculate the volume

V= a^3

V=4^3

V=64\ cm^3

The change temperature of the cube is

\Delta T=T_{f}-T_{i}

Put the value into the formula

\Delta T=67-19 = 48^{\circ}C

We need to calculate the increases volume

Using formula of increases volume

\Delta V=V\beta\Delta T

Put the value into the formula

\Delta V=64\times69\times10^{-6}\times48

\Delta V=0.212\ cm^3

Hence, The volume of an aluminum cube is 0.212 cm³.

5 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a
timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}

\omega \approx 17.781\,\frac{rad}{s}

Then, the tangential speed of the tack is: (\omega \approx 17.781\,\frac{rad}{s}, R = 0.393\,m)

v = \left(17.781\,\frac{rad}{s} \right)\cdot (0.393\,m)

v = 6.988\,\frac{m}{s}

The tangential speed of the tack is 6.988 meters per second.

7 0
3 years ago
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