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Nezavi [6.7K]
2 years ago
10

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.
Physics
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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Convert 5.5 kilometers into millimeters.​
dimaraw [331]

Answer:

5500000 millimeters

Explanation:

1 kilometre= 1000 meter

5.5 km=5.5 * 1000

=5500

Now,

1 metre = 1000 millimetres

5500 metre=1000*5500

=5500000 mm

4 0
2 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
hoa [83]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that  the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

5 0
3 years ago
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A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to
Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

f_{beat} = |f_2-f_1|

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

f_{beat} = |f_2-f_1|

5Hz = |f_2-440Hz|

f_2 = 440 \pm 5

f_2 = 445Hz or 435Hz

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

5 0
3 years ago
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