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Ymorist [56]
3 years ago
6

What is acceleration due to gravity

Physics
2 answers:
rjkz [21]3 years ago
7 0
The acceleration developed between two bodies containing mass as a result of the gravitational force between them.
Studentka2010 [4]3 years ago
5 0
If you stay on the same planet and drop a lot of objects one at a time,
it turns out that every object you drop falls from your hand to the ground
with the same acceleration, and hits the ground with the same speed,
no matter whether the object is light, heavy, or anything in between.

That particular value of acceleration is the "acceleration due to gravity".
On Earth, it's 9.81 meters per second².  On the moon, it's 1.62 meters
per second².  On Jupiter, it's 25.89 meters per second².

Why we don't generally notice it:  The previous description is true if the
ONLY force on the object is the force of gravity.  If it has to fall through
<u>air</u> on the way down, then the air can have a great effect on it.  Many
museums have an exhibit where they drop things in a long tube with
all the air removed from it, and there you can see some pretty weird
stuff ... like a bowling ball, a rock, a sheet of paper, and a feather, all
falling together, with nothing fluttering.

<u>Why</u> everything falls with the same acceleration ?  That's a separate question.




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8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an
77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = \frac{F}{\frac{\pi }{4}(d^2) }

To find force (F) =

                    120 = \frac{F}{\frac{\pi }{4}(100^{2} )}

                     F = 120 * (π/4) * (100)

                     F = 9425N

Calculate the true strain  (ε) = In (l₀ / l₁)

where

l₀ =  initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

  = In 1.1667

  = 0.154cm/cm

Calculate the true stress (σ) at fracture point

          = \frac{F}{\frac{\pi }{4}(d^2) }}

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = \frac{9425}{\frac{\pi }{4}(9.26^2) }}

   = 140MPa

To find the strength coefficient (K) of the material bar

K = \frac{140}{\sqrt{0.154} }

K = \frac{140}{0.3925}

   = 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K  = 356.75MPa

σ = 356.75(0.05)^0^.^5

  = 356.75 ( 0.2236)

  = 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0
3 years ago
A solid steel column is 4.0 m long and 0.20 m in diameter. Young's modulus for this steel is 2.0 × 1011 N/m2. By what distance d
zzz [600]

Answer:

3.12 x 10^-5 m

Explanation:

Length of steel column, L = 4 m

diameter, d = 0.2 m

radius = half of diameter = 0.1 m

Young's modulus, Y = 2 x 10^11 N/m^2

Mass of truck, m = 5000 kg

Force, F = mass of truck x acceleration due to gravity

F = 5000 x 9.8 = 49000 N

Area of crossection of cable,

A =  \pi r^{2}=3.14\times\0.1\times0.1=0.0314m^{2}

Let ΔL be the shrink in length of cable, then by the formula of Young's modulus

Y=\frac{F\times L}{A\times\Delta L}

\Delta L = \frac{F\times L}{A\times Y}

\Delta L = \frac{49000\times 4}{0.0314\times 2\times10^{11}}

ΔL = 3.12 x 10^-5 m

Thus, the shrink in the length of cable is 3.12 x 10^-5 m.

8 0
3 years ago
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