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2 years ago

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In the process of loading a ship, a shipping container gets dropped into the water and sinks to the bottom of the harbor. Salvag

e experts plan to recover the container by attaching a spherical balloon to the container and inflating it with air pumped down from the surface. The dimensions of the container are 5.80 m long, 2.60 m wide, and 2.80 m high. As the crew pumps air into the balloon, its spherical shape increases and when the radius is 1.50 m, the shipping container just begins to rise toward the surface. Determine the mass of the container. You may ignore the weight of the balloon and the air in the balloon. The density of seawater is 1027 kg/m3.

1 answer:

jolli1 [7]2 years ago

3 0

Answer:

57.885.8 kg weight of the container

Explanation:

The volume of the balloon * density of water = buoyant force of balloon

volume of a sphere = 4/3 pi r^3

= 4/3 pi * (1.5)^3 = 14.14 m^3 <===balloon volume

Now, find the buoyant force on the container ALONE ....

5.8 * 2.6 * 2.8 * 1027 = 43 364 kg <===== buoyant force

Now add the buoyant force of the balloon to find the weight

43 364 + 14.14 * 1027 = 57885.8 kg

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Nitella [24]

F = 52000 N

m = 1060 kg

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3 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

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3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

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(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

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From which we get

r = 1.32·R $_E$

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ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

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Answer:

See the attached pictures for detailed steps.

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Answer:

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