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photoshop1234 [79]
3 years ago
9

If an object is neutrally buoyant (does not sink or float) in fresh water, the same object placed into salt water wouldA sink.B

either sink or float.C do nothing.D float.
Physics
1 answer:
Grace [21]3 years ago
4 0

Answer:

D float.

Explanation:

Here, neutrally buoyant in fresh water

Now,

since the specific gravity of salt water is higher than the specific gravity of the fresh water therefore, the salt water will apply more buoyant force on the object.

And as the object is neutral in fresh water, more buoyant force will make the object float in the salt water.

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A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and
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Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

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I_2=2\ kg-m^2

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I_1\omega _1=I_2\omega _2

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A resistor resists the flow of electricity and usually converts electrical energy to heat energy. True or False?
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Read 2 more answers
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
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