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Trava [24]
3 years ago
9

A statue of unknown volume and density is suspended from a string. When suspended in air, the tension in the string is Tair; how

ever, when submerged in water, the tension is Tw

Physics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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All of the following are types of symbiotic relationships except for
nevsk [136]
The answer is D. the way I remember it is they all end with -alism.

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3 years ago
What is the trend noticed with the inner planets
lutik1710 [3]
The inner planets are closer to the Sun and are smaller and rockier. ... The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars.Apr 23, 2014
7 0
3 years ago
a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s.
8090 [49]
When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula
v^2 = u^2 - 2gh
where 
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.

Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m

Answer: 0.074 m (nearest thousandth)
5 0
3 years ago
Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free
sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

           ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

           ΔE = W

internal energy can be related to the change in temperature

            ΔE = 3/2 n K ΔT

we substitute

           3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

            W = 3/2 n (T₁- T₂)

3 0
3 years ago
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0
3 years ago
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