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3 years ago

What are some of the causes that disrupt the natural balance of the ecosystem?

Physics

2 answers:

dangina [55]3 years ago

5 0

Overpopulation of predators

Soloha48 [4]3 years ago

3 0

Answer: overpopulation of predators.

Explanation:

Overpopulation of predators is the correct option. Predators are the organisms which kill other organism so as to obtain them as food. The organism being killed is called as the prey.

If the population of the predators increase this will lead to the extinction of many species of prey. As it is evident that every organism in the ecosystem has a typical role to play. Therefore, it can be said that role of prey will not be performed and this may lead to disturbance in the ecological balance.

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9. A 12 v battery is connected to four 5 ohm light bulbs. What is the equivalent

vlabodo [156]

20 ohms 5 ohms
12volts

7 0

3 years ago

A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

$\eta = 1 - \frac{T_2}{T_1}$

Where,

$T_2 =$ Temperature of Cold Body

$T_1 =$ Temperature of Hot Body

$\eta$ = Efficiency

According to the statement our values are:

$\eta = 0.3$

$T_2 = 210K$

Replacing we have that

$0.3 = 1- \frac{210}{T_1}$

$\frac{210}{T_1} = 0.7$

$T_1 = \frac{210}{0.7}$

$T_1 = 300K$

Therefore the temperature of the heat source is 300K

6 0

3 years ago

An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor

MrMuchimi

The electric flux through the hole is $56.45\ webber$ .

Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
Mathematically it is given by $\phi_E=E.A \ Nm^2/C$ where E is the electric field and A is the area.
Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere $= 4\pi r^2$

$= 4\times 3.14\times (10 \times 10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2$

Area of the hole ( both side ) $= 2\times \pi r^2$

$= 2\times 3.14 \times (10^-^3)^2\\= 6.28 \times 10^-^6 m^2$

According to Gauss's theorem, the flow from a particular charge in the center is given by

$\phi= \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6$

This flux flows through the surface of the sphere, so the flux per unit area which is given by

$= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \ weber / m^2$

Flux through area of hole is given by :

$= 8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber$

Learn about more electric flux here :

brainly.com/question/26289097

#SPJ4

8 0

2 years ago

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

3 years ago

A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was

Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0

3 years ago