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1 year ago

A ball dropped from a window strikes the ground 2.71 seconds later. How high is the window above the ground?

Physics

1 answer:

koban [17]1 year ago

4 0

$s=36.52m$

Given ,

Time =2.71 second

We apply the formula

$s=ut+12at^2$

$u=0,t=2.73 \\a=g \\So, the \\height \\ s=36.52m$

When a body is lying on a horizontal surface, the object is subjected to a downward force due to the action of gravity. These gravitational forces always aim to attract the item, confirming the force's nature as an attractive force. Consider an object of mass "M" resting on a horizontal surface; the force exerted to the object by gravity is equal to the weight of the object "W." The acceleration owing to gravity, indicated as "g," is the acceleration by which an object is drawn towards the ground. As a result, Newton's second law states that when the values in the above connection are substituted.

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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.

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Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t

Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

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Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

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(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

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1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

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Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi

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1 year ago