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3 years ago

7

How is kinetic energy related to the distance

2 answers:

astra-53 [7]3 years ago

8 0

Kinetic energy =

1/2 (mass) • (velocity squared).

I don't see distance in that formula anywhere, so they're not related.

vivado [14]3 years ago

7 0

Answer:

ionk

Explanation:

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a sound pulse emitted underwater reflects off a school of fish and is detected at the same place 0.01 s later. how far away are

andrew-mc [135]

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3 years ago

A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg

ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

$Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3$

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

$p=m/V\\\\ p=m/(L*A)\\p*A=m/L$

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$\sqrt \frac{F*L}{m}\\$

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Substituting the values

$\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3} \\$

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4 years ago

What is Force!????!?

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3 years ago

Read 2 more answers

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

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liq [111]

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