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3 years ago

A solid object has a density of 2.85 g/cm3. According to Archimedes, what can be done to make this object float in water?

1 answer:

4 0

2: It's not just the capillary action, but the pull from transpiration (the evaporation of water from the tree) that is used to pull water up from the roots.

The second question needs context. Strong bonds alone won't cause tension. I don't see how adhesion is different. High vapour pressure could do it, but it's the difference in pressures that'd cause tension (and the resistance of that pressure by the surface). So, a low and high pressure would be needed. Poorly worded question :( 

1: "Adhesion is the tendency of certain dissimilar molecules to cling together due to attractive forces." [1] 

3: The other three answere would not work. Think of a boat. 

3: If you push gas, it will be compressed(get smaller). If you push liquid it will push something else. Thus, liquids are good for transferring force. This is a hydraulic system.

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A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be in a s

Greeley [361]

Answer:

$a = 4.76 cm$

Explanation:

As we know by the law of diffraction through circular aperture

$a sin\theta = 1.22 \lambda$

here we know that

a = diameter of the aperture

$\theta$ = diffraction angle

$\lambda$ = wavelength

now we have

$a = \frac{1.22 \lambda}{sin\theta}$

Now in order to find the wavelength we know

$\lambda = \frac{c}{f}$

$\lambda = \frac{343}{9100}$

$\lambda = 0.0376 m$

now we have

$a = \frac{1.22(0.0376)}{sin75}$

$a = \frac{0.046}{0.966}$

$a = 0.0476 m= 4.76 cm$

6 0

3 years ago

Name the nutrients required for the body.

Anettt [7]

Answer:

1- water

2- fat

3- carbohydrates

4- vitamins

5- minerals

6 0

3 years ago

Sleep is a difficult problem for dolphins because dolphins __________. A. need more sleep than their predators B. must be awake

Alex17521 [72]

Need more sleep than their predators

5 0

3 years ago

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Sauron [17]

I need a picture plz I don’t know what to answer.

6 0

3 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago