Look at the diagrams. Each model the arrangement of particles in a substance.

Atmospheric pressure varies from day to day. The level of a floating ship on a high-pressure day is (a) higher (b) lower, or (c)

frez [133]

Answer:

(c) no different than on a low-pressure day.

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.

<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>

5 0

2 years ago

A street light is on top of a 8 foot pole. Joe,

zubka84 [21]

8/4 = y/y-x

8y - 8x = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps

5 0

3 years ago

__ is the apparent in the position of an object when viewed from two locations

Yakvenalex [24]

The apparent shift <span>is the apparent in the position of an object when viewed from two locations.

Hope this helps :)</span>

3 0

2 years ago

A country would place a tariff on imported steel to

Whitepunk [10]

. . . 'protect' its domestic steel industry, by
increasing the price of imported steel.

7 0

3 years ago

Read 2 more answers

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

2 years ago