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hram777 [196]
3 years ago
15

(1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of

the clothesline, it sags a distance of 2 meters. What is the magnitude of the tension on the ends of the clothesline

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0

Answer:

Ft = 50.25 N

Explanation:

As you can see in the attached diagram, the angle θ is given by

tanθ = opp/adj

The opposite is the sag of distance 2 m and the adjacent is the half of the horizontal clothesline that is 20 m.

tanθ = 2/10

θ = tan⁻¹(2/10)

θ = 11.30°

There are two forces along the y axis, one is the F = mg acting downwards and the other is the Ftsinθ component of tension force in the string.

The sum of forces along y-axis is

5Ftsin(11.30) - mg = 0

Ft = mg/5sin(11.30)

Ft = 5*9.8/5sin(11.30)

Ft = 9.8/0.195

Ft = 50.25 N

Therefore, the magnitude of the tension on the ends of the clothesline is 50.25 N.

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How much does a 44 kg child weigh on Earth?​
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An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3
Tomtit [17]

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

v = u+at

t = \dfrac{v-u}{a}

Put the value into the formula

t = \dfrac{80-0}{3}

t =26.67\ sec

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times3\times(26.67)^2

s =1066.9\ m

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

7 0
2 years ago
How long with it take an object to accelerate from 6 m/s to 13 m/s at a rate of 1.4<br> m/s2. *
Blizzard [7]

Explanation:

hello,

a = ( v - u ) / t

where u is the initial velocity.

and v is the final velocity.

t represents time,

and a represents acceleration.

in this case,

a = 1.4 m/s²

u = 6 m/s

v = 13 m/s

hence,

1.4 = (13 - 6)/t

1.4t = 7

t = 7/1.4

t = 5 s

thank you!

7 0
2 years ago
A grating has 460 rulings/mm. What is the longest wavelength for which there is a 6.0th-order diffraction line
mestny [16]

Answer:

λ = 3.62 x 10⁻⁷ m = 362 nm

Explanation:

The grating equation gives the relationship between the wavelength, the diffraction line order and the diffraction angle. The grating equation is written as follows:

mλ = d Sinθ

where,

m = order of diffraction = 6

λ = longest wavelength = ?

d = 1/(460 rulings/mm)(1000 mm /1 m) = 2.17 x 10⁻⁶ m/ruling

θ = Diffraction angle = 90° (for longest wavelength)

(6)λ = (2.17 x 10⁻⁶ m/ruling) Sin 90°

λ = (2.17 x 10⁻⁶ m/rulings)/6

<u>λ = 3.62 x 10⁻⁷ m = 362 nm</u>

3 0
3 years ago
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