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3 years ago

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Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

their effective spring constant? 004393 Submit Answer Incorrect. Tries 4/99 Previous Tres If a mass of 0.31 kg is attached what will be the frequency of oscillation?

1 answer:

Llana [10]3 years ago

4 0

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

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Mass of the displaced material. In water it would be the mass of the water that the volume of the ball displaces.

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A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato

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Answer:

A) The speed of the potato at the lowest point of its motion is 7.004 m/s

B) The tension on the string at this point is 8.5347 N

Explanation:

Here we have that the height from which the potato is allowed to swing is 2.5 m

Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)

Where:

ω₂ = Final angular velocity

ω₁ = Initial angular velocity = 0 rad/s

α = Angular acceleration

θ₂ = Final angle position

θ₁ = Initial angle position

However, we have potential energy of the potato

= Mass m×Gravity g× Height h

= 0.29×9.81×2.5 = 7.1125 J

At he bottom of the swing, the potential energy will convert to kinetic energy as follows

K.E. = P.E. = 7.1125 J

1/2·m·v² = 7.1125 J

Therefore,

v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05

∴ v = √49.05 = 7.004 m/s

B) Here we have the tension given by

Tension T in the string = weight of potato + Radial force of motion

Weight of potato = mass of potato × gravity

Radial force of motion of potato = mass of potato × α,

where α = Angular acceleration = v²/r and r = length of the string

∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)

T = 8.5347 N

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3 years ago

Read 2 more answers

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

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(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

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Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

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A shopping cart given an initial velocity of 2.0 m/s north undergoes a constant acceleration of 3.0 m/s2 north. what is the dist

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