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3 years ago

11

A car is rounding a curve that is the arc of a circle with a radius of r . if the car has a constant speed of 10 m/s, and its ac

celeration is 5 m/s2 , what is the radius of the curve?

1 answer:

aev [14]3 years ago

4 0

The formula to find the radius is R=V^2/A or Radius=Velocity^2/Acceleration so that means
R=V^2/A
10^2/5=R
100/5=R
R=20m

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1. A huge spool of wire, 10,000 meters long, weighs 81.34 N. You cut off a meter or so and tie it between two posts, 0.660 m apa

WARRIOR [948]

Answer:

The beat frequency is 6.378 Hz.

Explanation:

Given that,

Length of wire = 10000 m

Weight = 81.34 N

Distance = 0.660 m

Tension = 52 N

Frequency = 196 Hz

We need to calculate the mass of the wire

Using formula of weight

$W= mg$

$m = \dfrac{W}{g}$

Put the value into the formula

$m=\dfrac{81.34}{9.8}$

$m=8.3\ kg$

The mass per unit length of the wire

$m_{l}=\dfrac{m}{l}$

Put the value into the formula

$m_{l}=\dfrac{8.3}{10000}$

$m_{l}=8.3\times10^{-4}\ kg/m$

We need to calculate the frequency in the wire

Using formula of frequency

$f_{w}=\dfrac{1}{2l}\sqrt{\dfrac{T}{m_{l}}}$

Put the value into the formula

$f_{w}=\dfrac{1}{2\times0.660 }\sqrt{\dfrac{52}{8.3\times10^{-4}}}$

$f_{w}=189.62\ Hz$

We need to calculate the beat frequency

Using formula of beat frequency

$f_{b}=f_{in}-f_{w}$

Put the value into the formula

$f_{b}=196-189.622$

$f_{b}=6.378\ Hz$

Hence, The beat frequency is 6.378 Hz.

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3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

or

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

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4 years ago

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