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elena55 [62]
2 years ago
5

On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

weight of the balloon, including its load and the hot air in it, is 20,000 N. a. Show that the weight of the displaced air is 20,000 N. b. Show that the volume of the displaced air is 1700 m3 .
Physics
1 answer:
Stella [2.4K]2 years ago
3 0

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

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kobusy [5.1K]

Answer:

Explanation:

a = 4ms⁻²,  Vf = 180 m/s  &  Vi = 140m/s

a = \frac{Vf-Vi}{t}

4 = \frac{180-140}{t}

t = 40/4

t = 10sec

To Measure Distance Use third Equation of Motion:

2aS = Vf²-Vi²

S = \frac{180*180 - 140*140}{2(4)}

S = 12800/8 = 1600m

4 0
3 years ago
Please answer D in the image with an explanation
puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0
2 years ago
Light traveling in air is incident on the surface of a block of plastic at an angle of
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Answer:

v = 2.51 x 10⁸ m/s

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v = c/n = 3.00e8/1.19 = 2.51e8 m/s

7 0
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A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0° below the horizontal. The coeffici
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Answer

given,

mass of the package = 12 kg

slides down distance = 2 m

angle of inclination = 53.0°

coefficient of kinetic friction = 0.4

a) work done on the package by friction is

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                   = -μk (mg cos 53°)(2.0)

                   =-(0.4)(8.0 )(9.8)(cos 53°)(2.0)

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b)

work done on the package by gravity is

             W_g = m (g sin 53°) d

                   = (8.0 )(9.8 )(sin 53°)(2.0 )

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c)

the work done on the package by the normal force is

             W_n = 0

d)

the net work done on the package is

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