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Pepsi [2]
3 years ago
6

Why does lightning hit taller objects?

Physics
1 answer:
AveGali [126]3 years ago
7 0

Answer:

because it has this rod thats metal on top of it so thats why it hits them.

Explanation:none

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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
4 years ago
Please help!!!! 70 points, i really need this! D:
jolli1 [7]

1. Frequency 2. measure from trough to trough

7 0
3 years ago
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A car travels east at 11 km/h for 3.1 hours and west at 11 km/h for 5.3 hours. What distance (in km) does it travel in this time
Verizon [17]

Answer:

45.1/3× 3.1/2 + 50.1/2× 2.1/2

=136/3. 7/2.+ 101/2× 5/2

=952/6 + 505/2

=( 952+ 1515)/6

=411.16

Explanation:

Make sure to read and go with it

4 0
3 years ago
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Hiii please help i’ll give brainliest if you give a correct answer please thanks!
MaRussiya [10]

Answer:

i think number 4 :/ i hope its right...

6 0
3 years ago
A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h
bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²        

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²    

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down  (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of  11 m for the first time.

b)  Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

5 0
4 years ago
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