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Natalija [7]
3 years ago
7

In which case will an object float on a fluid? A) Buoyant force is greater than weight. B) Buoyant force is less than weight. C)

Buoyant force equals weight. D) Buoyant force equals zero.
Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The correct option is A

Explanation:

Buoyancy can be described as the upward force that causes an object to float on water. <u>When the buoyant force of a liquid is greater than the weight of an object, the object will move to the surface of the liquid and float (because the buoyant force would be able to push it upwards)</u>. If the buoyant force is lesser than the weight of the object, the object will sink (because the buoyant force would not be able to push the object upwards). And when the buoyant force and weight of an object cancel out, the object would be suspended at the depth that this occurs.

Thus, from the explanation above, it can be deduced that for an object to float, buoyant force must be greater than weight of the object. Thus, the correct option is A.

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The volume of a gas depends upon its mass true or false
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3 years ago
(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing nort
Veronika [31]

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is D=\sqrt{A^{2} +B^{2} } and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as \frac{dA}{dt} and \frac{dB}{dt}. The rate of change of distance is also notated as \frac{dD}{dt}. Now, we have to find \frac{dD}{dt} from the Pythogorean theorem. If we derive the Pythogorean expression D=\sqrt{A^{2} +B^{2} } , we would have:

\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

\frac{dC}{dt} =\frac{1}{2}*(142^{2}   +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19) and the answer is 28.6 knots.

6 0
3 years ago
Edgar is walking 0.5 M/S toward the back of a train that is traveling forward at 6.0 M/S west. What is Edgars velocity relative
iragen [17]

Answer:

Just in 1 dimension  

Vpg = Vps + Vsg

Where

Vpg= relative to the ground

Vps= person relative to the ship

Vsg= ship relative to the ground

Vpg= Vps+Vsg ;

Vpg= 0.5 + 6  

Vpg= 6,5 m/s

8 0
4 years ago
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