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nekit [7.7K]
3 years ago
14

Misconception about falling objects

Physics
1 answer:
Yakvenalex [24]3 years ago
8 0
A simple rule to bear in mind is that all objects (regardless of their mass) experience the same acceleration when in a state of free fall. When the only force is gravity, the acceleration is the same value for all objects. On Earth, this acceleration value is 9.8 m/s/s.
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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
3 years ago
Which statements about acceleration are true?
bonufazy [111]
It's definitely not B or C. There are things missing from A and D so we can't narrow it down any farther.
4 0
3 years ago
What is the attractive force between all matter in the universe?
kvv77 [185]
The attractive force between all matter in the universe is gravity.
4 0
3 years ago
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What is the difference between heat exhaustion and heat stroke?
balu736 [363]
I believe the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.
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A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the
Nezavi [6.7K]

We use the equation of motion for vertical component,

s_{y} = u_{y} t+\frac{1}{2} gt^2.

Here, s_{y}   is displacement of bullet, u_{y}  is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

s_{y} =\frac{1}{2}gt^2

Given, s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m

Substituting the values, we get time of flight

2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s

4 0
3 years ago
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