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3 years ago

10

The chart below shows the density of seawater samples collected from the Gulf of Mexico by four different groups. Which group of

data is most reliable?

1 answer:

liberstina [14]3 years ago

4 0

Answer:

Group 2.

Explanation:

I chose this because I saw the numbers. I look at all the numbers, and then I looked at group 2. Group 2 has the most.

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A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

which is identical to the expression of the field inside the shell.

(d) $-12.3 C/m^2$

We said that at r = a, a charge of -q is induced. The induced charge density will be

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) $-1.9 C/m^2$

We said that at r = b, a charge of +q is induced. The induced charge density will be

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

3 0

3 years ago

Soil can best be described as the

mr Goodwill [35]

C. It is C because it really is loose covering.

7 0

3 years ago

Which tool can be used to separate white light into different colors? *

swat32

Answer:

prisms

Explanation:

8 0

3 years ago

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In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead

fiasKO [112]

Answer:

8.01e-22

Explanation:

7 0

3 years ago

Find the kinetic energy K of a satellite with mass m in a circular orbit with radius R. Express your answer in terms of m, M, G,

ratelena [41]

Answer:

$K=\frac{GmM}{2R}$

Explanation:

The kinetic energy is defined as:

$K=\frac{mv^2}{2}(1)$

Here, m is the object's mass and v its speed. In this case the speed of the satellite is the orbital speed, which is given by:

$v_{orb}=\sqrt\frac{GM}{R}(2)$

Here, G is the gravitational constant, M is the mass of the object that the satellite is orbiting and R is the radius of its circular orbit. Replacing (2) in (1):

$K=\frac{mv_{orb}^2}{2}\\K=\frac{m(\sqrt\frac{GM}{R})^2}{2}\\K=\frac{GmM}{2R}$

3 0

4 years ago