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NeTakaya
3 years ago
13

A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which

is 1524 meters away?
Physics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

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Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

To learn more about Columb's law refer to the link;

brainly.com/question/1616890

#SPJ1

7 0
2 years ago
A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter
tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

<em>F = Mg</em>

<em>⇒F = 84.8×9.81</em>

<em>⇒F = 831.888 N</em>

Area of the contact point

<em>A = πR²</em>

<em>⇒A = π0.006²</em>

<em>⇒A = π0.000036 m²</em>

Area of the four points is

<em>4A = 0.000144π m²</em>

Pressure

p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0
3 years ago
Which quantity does a light-year measure?<br>distance<br>speed<br>time<br>volume​
marissa [1.9K]

Answer:

Distance (i think)

8 0
3 years ago
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luda_lava [24]

After 4 half-lives, there was 1 gram.

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1 half-life earlier, after 2 half-lives, there must have been 4 grams.

1 half-life earlier, after 1 half-life, there must have been 8 grams.

1 half-life earlier, when it started decaying, there must have been 16 grams. <em>(B) </em>

6 0
3 years ago
The period of an oscillating particle is 32 s, and its amplitude is 15 cm. at t = 0, it is at its equilibrium position. find the
morpeh [17]

At t=0, the particle was at its equilibrium position. The time period is 32 seconds, so in 8 seconds, it will reach the extreme location once, and hence, in 8 seconds, it will cover a distance equivalent to its amplitude.

5 0
3 years ago
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