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3 years ago

14

If the change in velocity increases, what happens to the acceleration during the same time period?

2 answers:

4vir4ik [10]3 years ago

4 0

When the velocity is increasing the acceleration increases too

maksim [4K]3 years ago

4 0

When the velocity is increasing the acceleration increases tooo

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What can happen when two plates bordering each other suddenly move and release energy

Ostrovityanka [42]

The correct answer for this question would be an EARTHQUAKE. What happens when two plates bordering each other suddenly move and release energy is an earthquake. When these plates slip one another and energy is released, this results in an earthquake. Hope this answer helps.

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3 years ago

How are the three lines of defense the same

xz_007 [3.2K]

Here is an example of how the three lines of defense are the same:

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3 years ago

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A gymnast does a one-arm handstand. The humerus, which is the upper arm bone between the elbow and the shoulder joint, may be ap

igor_vitrenko [27]

Answer:

$= 5.241 \times 10^{-5} m$

Explanation:

Given:

Length of cylinder is, L = 0.27 m

Outer radius of cylinder is, r_out = 1.12×10^{-2} m

Inner radius of cylinder is, r_in = 3.9×10^{-3} m

Mass of person, m = 60 kg

Young's modulus , Y = 9.4×10^9 N/m2

(a)

Compressional strain of humerous is,

$Strain = \frac{Stress}{Young's\ modulus}$

$\frac{\Delta L}{L_0} = \frac{\frac{F}{A}}{Y}$

$= \frac{(mg)}{\pi(r_out^2 - r_in^2 )Y}$

$= \frac{(60 kg)(9.8 m/s2 )}{(\pi)[( 1.12\times 10^{-2})^2 - (3.9\times 10^{-3} m)^2] (9.4\times 10^9 N/m2 )}[tex] [tex]= 1.80\times 10^{-4} m$

(b) Let assume that humerous is compressed by ΔL

Since, strain = ΔL/L0

$(1.80 \times 10^{-4} m) = ΔL / 0.29 m$

Thus,

$ΔL = (4.56 \times 10^{-4} m)(0.29 m)$

$= 5.241 \times 10^{-5} m$

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3 years ago

Gravity on earth is 9.8m/s^2, and gravity on the moon is 1.6 m/s^2. So if the mass of an object on earth is 40 kilograms, it’s m

tigry1 [53]

31.6 kilograms mass on the

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4 years ago

A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the osc

ludmilkaskok [199]

Answer:

.487 s⁻¹

Explanation:

Let damping constant be τ . The equation of decreasing amplitude can be written as

A = A₀ $e^{-\tau t$

A / A₀ = $e^{-\tau t$

At t = 9.43 s , A / A₀ = .01

.01 = $[e^{-\tau\times9.43$

ln.01 = - 9.43 τ

-4.6 = -9.43τ

τ = .487 s⁻¹

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4 years ago

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