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3 years ago

5

A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.Wh

at is the spring constant k?How long is the spring when a 4.0 kg mass is suspended from it?

1 answer:

Dmitrij [34]3 years ago

7 0

As the spring is stretched, it exerts an upward restoring force <em>f</em>. At maximum extension, Newton's second law gives

∑ <em>F</em> = <em>f</em> - <em>mg</em> = 0 ==> <em>f</em> = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke's law, if <em>k</em> is the spring constant, then

<em>f</em> = <em>kx</em> ==> <em>k</em> = <em>f</em>/<em>x</em> = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

<em>f</em> = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance <em>x</em> such that

39.2 N = (130 N/m) <em>x</em> ==> <em>x</em> = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

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awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall

pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

$KE = 6.66 J$

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

$m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}$

now plug in all values in it

$1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22$

so we have

$3.01 = 0.375v_1 + 0.927v_2$

similarly in Y direction we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}$

now plug in all values in it

$0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22$

so we have

$0 = 0.927v_1 - 0.375v_2$

$v_2 = 2.47 v_1$

now from 1st equation we have

$3.01 = 0.375 v_1 + 0.927(2.47 v_1)$

$v_1 = 1.13 m/s$

$v_2 = 2.78 m/s$

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

$KE = \frac{1}{2}(1.47)(3.01^2)$

$KE = 6.66 J$

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4 years ago

What is happening to the velocity of an object if it has a negative acceleration?

oksano4ka [1.4K]

Answer:

<h3>According to our principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object has a negative acceleration. In Example D,<u> the object is moving in the negative direction</u> (i.e., has a negative velocity) and is speeding up.</h3>

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3 years ago

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Select all of the statements that are true.

natta225 [31]

The correct statements are "Each orbit holds a fixed number of electrons" and "The n=1 orbit can only hold two electrons." According to the Bohr model, the maximum number of electrons that can occupy an orbit is given by $2n^2$ , where n is the number of the orbit. For instance, when n=1 it means $2n^2= 2(1)^2=2$ . This particular orbit can only hold up to two electrons. Even though the electrons can gain energy and move to higher orbits or electrons from higher orbits can lose energy and drop to the n=1 level, the energy level would not allow more electrons to enter the orbit once it is full. Again the octet rule, which states that atoms achieve stability by having 8 valence electrons, limits the maximum number of electrons that can be occupied by an orbit. The gain and loss of electrons is done to achieve the noble gas configuration and once that is reached no more electron can be added to an orbit

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3 years ago

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dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

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$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

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$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

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$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

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Aleks04 [339]

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