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3 years ago

14

Think of an animal and list two resources that it might compete for in its community . Then describe what adaptions the animals

has to compete for these resources

1 answer:

Viefleur [7K]3 years ago

6 0

An animal is a lion, it may compete for food and it’s territory. The animal may have to fight other animals to get these things

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What structure on the south african coast has a range of 63 km and releases flashes every 30 seconds?

I am Lyosha [343]

Answer:

Cape Point lighthouse

Explanation:

The lighthouse is located South of Cape Town, at a point where boats turn to go around the tip of Africa.

The new lighthouse has the most powerful light of all the lighthouses in South Africa, with a power of 10 megacandelas. That's why it's being visible for up to 63 km around.

It's a very important lighthouse helping to conduct the traffic around dangerous waters where 2 oceans meet.

7 0

3 years ago

The law of conservation of matter states that during a chemical reaction

Sveta_85 [38]

This law states that, despite chemical reactions or physical transformations, mass is conserved — that is, it cannot be created or destroyed — within an isolated system

4 0

3 years ago

If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.

Alex Ar [27]

The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

Hope it helped!

5 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

<u>r < a:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

<u>r = a:</u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>a < r < b:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>r = b:</u>

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

<u>r < c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u>At r = b:</u>

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>At r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

What is the connection between electrons and protons

TEA [102]

Answer:

Electrons and protons are connected with one another in term of charge and size.

Explanation:

The size of electron and proton is always same in any atom but they possess opposite charge.
Electron in any atom carries negative charge where as proton carries the positive charge.
In any neutral atom the charge between electron and proton is balanced along with the size.
The nucleus of any atom bounds only proton and neutron but the electron is present revolving around the nucleus.

7 0

3 years ago