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2 years ago

Can someone help asap plz don’t understand how to do this

Physics

1 answer:

shtirl [24]2 years ago

3 0

Homie I don’t know either‍♀️Drop out of schl ig ;-;

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Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

bearhunter [10]

Answer:

$strain = 1.4 \times 10^{-3}$

Explanation:

As we know by the formula of elasticity that

$E = \frac{stress}{strain}$

now we have

$E = 110 GPA$

$E = 110 \times 10^9 Pa$

Area = 15.2 mm x 19.1 mm

$A = 290.3 \times 10^{-6}$

now we also know that force is given as

$F = 44500 N$

here we have

stress = Force / Area

$stress = \frac{44500}{290.3 \times 10^{-6}}$

$stress = 1.53 \times 10^8 N/m^2$

now from above formula we have

$strain = \frac{stress}{E}$

$strain = \frac{1.53 \times 10^8}{110 \times 10^9}$

$strain = 1.4 \times 10^{-3}$

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2 years ago

What is the most commmon injury in the United states

Bess [88]

Home injury’s or at home falls

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2 years ago

Which kind of cycle is the rock cycle-one with a set path or one with many possible paths? Construct an explanation that describ

hjlf

Answer:

Every rock is somewhere in the rock cycle. The story of the rock is the story of rock cycle: where the rock is now on the rock cycle, the path it took on the rock cycle to get to where it is now, and the path on the rock

Explanation:

This is the answer i put it and it worked if it didn't work for you that's because you lesson or teacher was doing it in a different way.

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2 years ago

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Suppose that the ultrasound source placed on the mother's abdomen produces sound at a frequency 2 MHz (a megahertz is 10^610 6

ella [17]

Answer:

the maximum frequency observed is 2.0044 10⁶ Hz

Explanation:

This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is

f ’= $f_o \ ( \frac{v \pm v_o}{v} )$

the + sign is used when the observer approaches the source

typical speeds of a baby's heart stop are around 200 m / min

let's reduce to SI units

v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s

let's calculate

f ’= 2 10⁶ ( $\frac{1500 \ \pm 3.33}{1500}$ )

f ’= 2.0044 10⁶ Hz

f ’= 1,9956 10⁶ Hz

therefore the maximum frequency observed is 2.0044 10⁶ Hz

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2 years ago

Which gives the frequency of the wave in terms of its peaks? A. the number of peaks passing a point per second B. the height of

ELEN [110]

The answer is A. The frequency of a wave is how many waves pass a point (measured in this case by how many peaks pass) in an amount of time (in this case 1 second)

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