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andriy [413]
3 years ago
5

A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is

700 kg, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 2900N . The tension in the towrope between the transport plane and the first glider is not to exceed 12000 N.
A- If a speed of 40 m/s is required for takeoff, what minimum length of runway is needed?

B- What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?

Express your answer using two significant figures.
Physics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

a) Δx = 180.59 m

b) T = 6001 N

Explanation:

a)

According to Newton's second law, which says that acceleration is directly proportional to the net force, the equation is equal to:

ΣF = m*a = T-f

Clearing a, and solving:

a = (T-f)/m = (T-f)/2*m = (12000-5800)/(2*700) = 4.43 m/s^2

To evaluate the final speed the following equation will be used:

vf^2 = vi^2 + 2*a*Δx = 0 + 2*a*Δx = 2*a*Δx

Clearing Δx:

Δx = vf^2/2*a = (40 m/s)^2/(2* 4.43 m/s^2) = 180.59 m

b)

The tension is equal to:

T = m*a + f = (700 kg * 4.43 m/s^2) + 2900 N = 6001 N

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Answer:

Option D - 0.2 s

Explanation:

We are given;

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Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

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Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

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kati45 [8]

Answer:

12N to the right.

Explanation:

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Assign a negative value to forces towards the left, and a positive value to the forces towards the right: -3N and +15N

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3 years ago
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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Gap = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Potential difference is given by

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference between the plates is 10000 V

Area is given by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of the plate is 0.00225988700565 m²

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

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3 years ago
A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?
dedylja [7]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2.

Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

7 0
3 years ago
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