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2 years ago

8. Before leaving the ground an airplane traveling with constant acceleration makes a run on the

Physics

2 answers:

Effectus [21]2 years ago

8 0

Answer:

$\color{Blue}\huge\boxed{Answer}$

<h3>B. Speed at which it leaves the ground</h3>

Andreyy892 years ago

5 0

Answer:

Explanation:

a) s = ½at²

a = 2s/t² = 2(1800)/12² = 25 m/s²

b) v = at = 25(12) = 300 m/s

c) first second s = ½(25)1² = 12.5 m

twelfth second s = ½(25)(12² - 11²) = 287.5 m

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How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?

matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be $f_{1}$

So, paper will attain terminal velocity when mg = $f_{1}$

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = $f_{2}$

Rock will attain terminal velocity when Mg = $f_{2}$

Mg > mg

so, $f_{2}$ > $f_{1}$

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.

5 0

3 years ago

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0

2 years ago

How did you manage to overcome the difficulties that arose?

Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

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2 years ago

What is your least favorite candy? Explain

aev [14]

Answer:

My least favorite is whoppers.....Trust me i love chocolate, but not when it taste like chalk......

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2 years ago

Read 2 more answers

Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Natasha2012 [34]

The work done by $\vec F$ along the given path C from A to B is given by the line integral,

$\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r$

I assume the path itself is a line segment, which can be parameterized by

$\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath$

with 0 ≤ t ≤ 1. Then the work performed by F along C is

$\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}$

7 0

2 years ago