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3 years ago

6

Please help...........

1 answer:

oksian1 [2.3K]3 years ago

6 0

Answer:

Explanation:

tha question is too hard simplfy it fast

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A steel wire in a piano has a length of 0.680 m and a mass of 4.600 ✕ 10⁻³ kg. To what tension must this wire be stretched so th

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Answer:

728 N

Explanation:

$L$ = length of the wire = 0.680 m

$m$ = mass of the steel wire = 0.0046 kg

$f$ = Fundamental frequency = 261.6 Hz

$T$ = tension force in the steel wire

Fundamental frequency in wire is given as

$f = \frac{1}{2L} \sqrt{\frac{TL}{m} } \\261.6 = \frac{1}{2(0.680)} \sqrt{\frac{T(0.80)}{0.0046} }\\(2) (0.680) (261.6) = \sqrt{\frac{T(0.80)}{0.0046} }\\355.8 = \sqrt{\frac{T(0.80)}{0.0046} }\\355.8^{2} = \frac{T(0.80)}{0.0046}\\T = \frac{(355.8^{2}(0.0046))}{0.80} \\T = 728 N$

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A bus took 8 hours to travel 639 km. For the first 5 hours, it

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A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago