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3 years ago

I) Give two practicalapplication of static friction

Physics

1 answer:

bulgar [2K]3 years ago

7 0

1). Walking / Driving

If there were no static friction between the soles of your shoes and the ground, then you could move your feet back and forth but your body would never go anywhere.

Same for using tires to move a car, a bus, a bicycle or a motorcycle.

2). Sleeping

If there were no static friction between your jammies and the sheet, you would slide right off of the bed whenever there was the slightest breeze of air in the room.

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A crate falls from an airplane flying horizontally at an altitude of 2,000 m.

Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

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4 years ago

12345 [234]

The similarity is that they both are types of bonds in molecules.
Ionic bonds are between a metal and a nonmetal.
Covalent bonds are between two nonmetals.

8 0

3 years ago

Question 1 (1 point) Question 1 Unsaved

DochEvi [55]

Answer:

(1) fossel fuel

(2) Kyoto protocol

(3) all of the above

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3 0

3 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

Select the correct answer.

mrs_skeptik [129]

Answer:

Explanation:

net force = right force - left force

net force = 22 - 13=9N

4 0

3 years ago

I) Give two practicalapplication of static friction​

I) Give two practicalapplication of static friction