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VikaD [51]
4 years ago
10

A string of length 1.3 m is oscillating in a standing wave pattern. If the tension in the string is 430 N, the string has a mass

of 23 g/m, and the amplitude of the oscillations is 2.1 mm, what is the maximum speed of a point on the string when it is oscillating in the fundamental mode?
A

1.4 m/s

B

0.69 m/s

C

0.45 m/s

D

0.22 m/s

E

2.8 m/s
Physics
1 answer:
Vlad1618 [11]4 years ago
6 0

To solve this problem we will use the concepts related to the speed of a string which is given by the applied voltage and the linear mass density of it. With the speed value we can find the fundamental frequency that will serve as a step to find the maximum speed through the relation of Amplitude and Angular Speed. So:

v = \sqrt{\frac{T}{\mu_e}}

Where,

T = Tension

\mu_e= Linear mass density

v = \sqrt{\frac{430}{0.023}}

v = 136.7m/s

With this value the fundamental frequency would be

f = \frac{v}{2L}

f = \frac{136.7}{2*1.3}

f = 52.6Hz

Finally the maximum speed is given with the relation between the Amplitude (A) and the Angular frequency, then

V_{max} = A\omega

V_{max} = A(2\pi f)

V_{max} = (2.1*10^{-3})(2\pi 52.6)

V_{max} = 0.69m/s

Therefore the correct answer is B.

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777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

K = \frac{1}{2}mv^{2}

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

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3 years ago
A solution has a pH of 8. Which best describes the solution?
harkovskaia [24]

Answer:

D

Explanation:

a weak base is the answer

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3 years ago
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Whats mechanical energy mean
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Mechanical energy<span> is the sum of kinetic and potential </span>energy<span> in an object that is used to do work. In other words, it is </span>energy<span> in an object due to its motion or position, or both.

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A truck is parked at the bottom of a hill. It moves from the bottom of a hill
Marat540 [252]

Answer:

C

Explanation:

potential energy is stored energy that an object has due to its position or chemical composition. Since the truck has the possibility to move down the hill when it is at the top of the hill, it has a great amount of potential energy, howver when its at the bottom of the hill, it doesn’t have the possibility to move or can move v little due to its positions,  therefore it has little to no potienatal energy.

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A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m lon
ch4aika [34]

To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is

KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f

KE_i = 0, since initially cylinder was at rest

PE_f = 0 since at the ground potential energy is zero

The mathematical values are,

mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2

Here,

m = mass

g= Gravity

h = Height

V = Velocity

I = \frac{mr^2}{2} moment of Inertia in terms of its mass and radius

\omega = \frac{V}{r} Angular velocity in terms of tangential velocity and its radius

Replacing the values we have that

mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2

gh = \frac{v^2}{2}+\frac{v^2}{4}

v = \sqrt{\frac{4gh}{3}}

From trigonometry the vertical height of inclined plane is the length of this plane for sin\theta, then

h = 2.00*sin 25

h = 0.845 m

Replacing,

v = \sqrt{\frac{4(9.8)(0.845)}{3}}

V = 3.32 m/s

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s

6 0
3 years ago
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