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Andru [333]
3 years ago
7

One end of a string 6.26 m long is moved up and down with simple harmonic motion at a frequency of 95 Hz . The waves reach the o

ther end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm.
Physics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

Explanation:

Given

length of string L=6.26\ m

frequency f=95\ Hz

time taken by wave to reach at other end t=0.5\ s

speed of  wave is given by

v=\frac{length\ of\ string}{time\ taken}

v=\frac{6.26}{0.5}

v=12.52\ m/s

wavelength of is given by

\lambda =\frac{velocity}{frequency}

\lambda =\frac{12.52}{95}

\lambda =0.131\ m                          

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Answer: False

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When does resonance occur?
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4 years ago
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A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 22.0 N to keep this 7.00 kg
xxMikexx [17]

Answer:

B . 68.7 N

Explanation:

Given in the question that;

Force = 22 N

Mass = 7.0 kg

Velocity = 4.0 m/s

Gravitational force is the weight of wooden block , calculated as mass times acceleration due to gravity.

F= m* a

F= 7 * 9.81

F= 68.67 N

F= 68.7 N

7 0
3 years ago
Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler
ehidna [41]

A) Acceleration in part A: 6.1 m/s^2

B) Acceleration in part B: 2.7 m/s^2

C) Acceleration in part C: 1.5 m/s^2

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

\Delta t = 0.9 - 0 = 0.9 s

So the acceleration is

a=\frac{5.5-0}{0.9}=6.1 m/s^2

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

\Delta t = 2.4 - 0.9 = 1.5 s is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

a=\frac{9.5-5.5}{1.5}=2.7 m/s^2

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

\Delta t = 3.4 - 2.4 = 1 s is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

a=\frac{11-9.5}{1}=1.5 m/s^2

Learn more about acceleration:

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3 0
3 years ago
A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static fricti
vfiekz [6]

Answer:

0.255

Explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

7 0
3 years ago
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