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valentina_108 [34]
3 years ago
5

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

ir is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Aleks [24]3 years ago
3 0

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

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A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed
IgorC [24]

Answer:

The direction and magnitude of velocity is 38.65° and 12.005 m/s

Explanation:

Given that,

Mass of car = 1500 kg

Speed of car = 25.0 m/s

Mass of van = 2500 kg

Speed of van =20.0 m/s

We need to calculate the velocity

Using conservation of energy

m_{c}u_{i}+m_{v}u_{i}=(m_{c}+m_{v})v_{f}

1500(25i+0j)+2500(0+20j)=4000(v_{f})

v_{f}=\dfrac{1500\times25}{4000}i+\dfrac{1500\times20}{4000}j

v_{f}=9.375 i+7.5 j

The magnitude of velocity

|v_{f}|=\sqrt{(9.375)^2+(7.5)^2}

|v_{f}|=12.005\ m/s

We need to calculate the direction

\tan\theta=\dfrac{coefficient\ of\ j}{coefficient\ of\ i}

\tan\theta=\dfrac{7.5}{9.375}

\theta=\tan^{-1}0.8

\theta=38.65^{\circ}

Hence, The direction and magnitude of velocity is 38.65° and 12.005 m/s.

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All of the following show friction as a useful force, except
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having to push a rough and heavy box across the floor to move it

Explanation:

Friction is not a useful force because we have to exert even more force to push a body that is rough and heavy across the floor to move it.

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Friction is useful in that:

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The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank
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When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

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now by above force balance equation we can write

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