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dexar [7]
3 years ago
7

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 1900 N. The tension in the towrope between the transport plane and the first glider is not to exceed 12000 N. Part A If a speed of 40 is required for takeoff, what minimum length of runway is needed? Express your answer using two significant figures. Part 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. Please try to explain how you get to the answer.
Physics
1 answer:
velikii [3]3 years ago
8 0

Answer:

(a)  Length =136.58 m

(b)  T=5995 N

Explanation:

for the glider in the back

T - 1900 = 700 a

for the glider in front

12000-T -1900 = 700a

add equations

12000-3800 = 1400 a

a=5.85 m/s^2

v^2 = v0^2 + 2 a x

40^2 = 2*5.85*x

Length =136.58 m

b) plug the a back into one of the previous formula

T -  1900 = 700*5.85

T=5995 N

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Explanation:

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7 0
3 years ago
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0
3 years ago
Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea
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3 0
4 years ago
Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The fir
Ivan

Answer:

The distance s of how far the ball will go at the highest setting = 2.25m

Explanation:

Let consider x to be the representative of the compression and the distance to be s

Recall that:

\dfrac{1}{2}\times K \times  x^2 = mgs +c

By cross multiplying

K \times  x^2 = 2(mgs +c)

K \times  x^2 = 2\times 9.81(ms) +2c

K \times  x^2 = 19.62(ms) +2c

x^2 = A \times  s+B

Thus, for the first setting

x = 1 , s = 0.25

for the second setting

x = 2,   s = 1

1 = 0.25A + B ---  (1)

4 = A + B    ----- (2)

From (1); let B =  1 - 0.25A  and substitute it into (2)

4 = A + 1 - 0.25 A

4 - 1 = A - 0.25 A

3 = 0.75 A

A = 3/0.75

A = 4

From (2)

4 = A + B

4 = 4 + B

B = 4 - 4

B = 0

Therefore, for the highest setting, where x = 3

Then :

x^2 = A \times  s+B will be:

3² =   4s + 0

9 = 4s

s = 9/4

s = 2.25 m

∴

The distance s of how far the ball will go at the highest setting = 2.25m

6 0
4 years ago
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