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kherson [118]
3 years ago
7

For a submarine, identify when it will have the GREATEST buoyant force acting on it. Assume that neither the volume of the subma

rine nor the density of the water change. Choose all that apply.
a. When it is empty and fully submerged at a depth of 89 m.
b. When it is empty, fully submerged, and resting on the bottom at a depth of 178 m.
c. When it is in dry dock and no part of it is submerged in water.
d. When it is full of cargo and floating on the surface.
e. When it is empty and floating on the surface.
f. When it is full of cargo, fully submerged and resting on bottom at a depth of 178 m.
g. When it is full of cargo and fully submerged at a depth of 89 m.
Physics
1 answer:
lidiya [134]3 years ago
6 0

Answer: a, c, and g

Explanation:

Buoyant Force is an upward force acting on submerged object equal to weight of fluid displaced by the submerged object.

If no part is submerged (V = 0) that is volume. Therefore there is Zero Buoyant Force.

Fully submerged produces greatest buoyant force since greatest amount of fluid was displaced.

Whenever it is fully submerged it will have the greatest buoyant force.

Buoyant Force DOES NOT Depend on Depth

A fully submerged object displaces its volume in fluid

A floating object displaces its weight in fluid.

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Through refraction , it bends as it passes into a solid object
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A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin
pychu [463]

a) See graph in attachment

b) The suvat equation to use is v_f^2 - v_i^2 = 2as

c) The acceleration is a=\frac{v_f^2-v_i^2}{2s}

d) The acceleration is 1.25 m/s^2

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is x_i = 0, the origin

- The  final position of the boat is x_f = 200 m

- The initial velocity of the boat is v_i = 20.0 m/s

- The final velocity of the boat is v_f = 30.0 m/s

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

v_f^2 - v_i^2 = 2as

where

a is the acceleration

c)

Now we have to solve the equation

v_f^2 - v_i^2 = 2as

In order to find the acceleration.

This can be done by dividing both terms by 2s: this way, we find

\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}

And so the acceleration is

a=\frac{v_f^2-v_i^2}{2s}

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

And substituting into the equation,

a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

v_f = v_i + at

where:

v_i is the initial velocity

v_f is the final velocity

a is the acceleration

t is the time

Here we have:

v_i = 20.0 m/s

v_f = 30.0 m/s

a=1.25 m/s^2

Solving for t, we find:

t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s

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Answer:

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ξ  = 2Nπr²B/dt

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ξ = 44 × π × 0.2601  × 0.000047/0.2

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