How does the ecotourism industry act as an ecosystem service that benefits tourists, the people that live in the area, and the people that work in the industry?

Answers

Answer 1

It promotes enjoyable experiences for both guests and hosts; It lessens the negative effects of tourism on the environment; It increases local residents' access to work and financial prospects; It promotes conservation by offering financial advantages in its direction.

What is ecotourism?

The World Tourism Organization defines ecotourism as all forms of nature-based travel in which visitors' primary interests are in observing and appreciating the environment as well as the local cultures that are still practiced in natural settings.

Here are a few instances of ecotourism:

Visit places where there is nature (think of the flora, animals, and resources)discovering the heritage of the local culture.Ecologically sound wilderness excursions.volunteering or watching nature.trips devoted to environmental preservation efforts, like beach clean-ups.

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Related Questions

Why did it matter that tropical plant fossils were found in Antarctica?
A. It supported the idea that the southern hemisphere was once in the north and vice versa. B. It supported the idea that there has been climate change in the past.
C. It supported the idea that the continents used to be connected and then drifted.
D. It supported the idea that tropical plants can actually grow in cold climates.
E. It supported the idea that people had migrated from the tropics to what is now known as Antarctica.

Answers

The correct option is

C. It supported the idea that the continents used to be connected and then drifted.

The disclosure of tropical plant fossils in Antarctica was noteworthy since it gave proof for the hypothesis of a mainland float, which recommends that the landmasses were once associated with a supercontinent called Pangaea and have since moved to their current positions.

The nearness of tropical plant fossils in Antarctica proposes that the landmass was once found in a hotter climate closer to the equator, and has since floated to its current area close to the South Shaft.

This revelation backed the thought of plate tectonics and mainland float, which has gotten to be broadly acknowledged within the logical community. In this manner, choice C is the right reply. 

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help what’s the answer for this

Answers

The acceleration of mass, m2 is determined as 4.2 m/s².

What is the acceleration m2?

The acceleration of mass, m2 is calculated by applying Newton's second law of motion, which states that force applied to a body is proportional to the product of mass and acceleration of the body.

F(net) = ma

m2g - m1g = m2a + m1a

a = (m2g - m1g)/(m2 + m1)

where;

g is acceleration due to gravitya is the acceleration of the blocks

a = (5 x 9.8  -  2 x 9.8)/(5 + 2)

a = 4.2 m/s²

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Ammeters that are all placed in series should have _____ reading anywhere in the circuit.
A. A different
B. Random
C. The same
D. Various

Answers

Ammeters that are all placed in series should have _The same____ reading anywhere in the circuit. The correct option is C

What is ammeter ?

An ammeter is an instrument used to measure the electric current in a circuit. It is connected in series with the circuit element or the part of the circuit in which the current is to be measured.

The ammeter is designed to have a low resistance, so that it does not significantly affect the current being measured. The current passing through the ammeter creates a magnetic field, which deflects a pointer on a scale or produces a digital readout.

Therefore,  Ammeters are commonly used in both AC and DC circuits to measure current flow.

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What is the period, in seconds, of this mass?

Answers

The period (in second) of the 0.95 Kg mass, given that it has an amplitude of 0.21 m and an angular velocity of 9.5 rad/s is 0.66 second

How do i determine the period of the mass?

From the question given above, the following data were obtained:

Mass (m) = 0.95 KgAmplitude (A) = 0.21 mAngular frequency (ω) = 9.5 rad/sPi (π) = 3.14Period (T) =?

The period of the mass can be obtained as shown below:

ω = 2π/ T

9.5 = (2 × 3.14) / T

9.5 = 6.28 / T

Cross multiply

9.5 × T = 6.28

Divide both sides by 9.5

T = 6.28 / 9.5

T = 0.66 second

Thus, from the above calculation, we can conclude that the period of the mass is 0.66 second

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What voltage will send a current of 5 amperes through a bell circuit if the resistance of the circuit is 4 ohms?

Answers

Answer:

Therefore, the voltage required is 20 volts

Explanation:

To find the voltage that will send a current of 5 amperes through a bell circuit with a resistance of 4 ohms, we can plug in the values into the equation and solve for V: V=IR

V=(5A)(4Ω)

V=20V

A 70.0 cm, uniform, 40.0 N shelf is supported horizontally by two vertical wires attached to the sloping ceiling (Figure 1). A very small 15.0 N tool is placed on the shelf midway between the points where the wires are attached to it.
a)Find the tension in the left-hand wire.
b)Find the tension in the right-hand wire.

Answers

Answer:

To solve this problem, we need to draw a free-body diagram of the shelf and apply Newton’s second law of motion to find the tension in each wire.

The free-body diagram of the shelf is shown below:

The forces acting on the shelf are its weight (W) and the forces exerted by the wires (T1 and T2). Since the shelf is at rest, the net force acting on it must be zero.

Using trigonometry, we can find that the angle between the ceiling and each wire is 30 degrees.

Now we can write two equations using Newton’s second law of motion:

ΣF_x = 0: T1 cos(30) - T2 cos(30) = 0

ΣF_y = 0: T1 sin(30) + T2 sin(30) - W - F_tool = 0

where F_tool is the force exerted by the tool on the shelf.

Substituting W = mg and F_tool = 15 N, we get:

T1 cos(30) - T2 cos(30) = 0

T1 sin(30) + T2 sin(30) = mg + F_tool

Solving these equations simultaneously gives:

T1 = (mg + F_tool) / (2 sin(30) cos(30)) ≈ 68.7 N

T2 = (mg + F_tool) / (2 sin(30) cos(30)) ≈ 51.5 N

Therefore, the tension in the left-hand wire is approximately 68.7 N and in the right-hand wire is approximately 51.5 N.

I hope this helps! Let me know if you have any other questions.

Explanation:

A hard-boiled egg of mass 50.0 g moves on the end of a spring with spring constant

Answers

The magnitude of the damping coefficient is = 0.277 Ns/m

The magnitude of the damping coefficient

We can use the formula for the damped harmonic motion of a spring-mass system:

x = A*e^(-bt/2m)*cos(wt + delta)

At t=0, the displacement of the egg is x = 0.300 m, and at t=5.00 s, the displacement has decreased to x = 0.100 m.

The angular frequency of the motion is:

w = sqrt(k/m) = sqrt(25.0 N/m / 0.0500 kg) = 10.0 rad/s

The equation for the amplitude of the motion at time t is:

A = x / cos(wt + delta) * e^(bt/2m)

At t=0, we have:

A = 0.300 m / cos(0 + delta) * e^(b0/20.0500 kg) = 0.300 m / cos(delta)

At t=5.00 s, we have:

A = 0.100 m / cos(10.0 rad/s * 5.00 s + delta) * e^(b5.00 s/20.0500 kg)

Dividing these two equations, we get:

0.300 m / cos(delta) / (0.100 m / cos(10.0 rad/s * 5.00 s + delta) * e^(b5.00 s/20.0500 kg)) = e^(b5.00 s/20.0500 kg)

Simplifying, we get:

cos(10.0 rad/s * 5.00 s + delta) * e^(b5.00 s/20.0500 kg) / cos(delta) = 3.00

Taking the natural logarithm of both sides, we get:

ln(cos(10.0 rad/s * 5.00 s + delta) * e^(b5.00 s/20.0500 kg) / cos(delta)) = ln(3.00)

Using the properties of logarithms, we can simplify this to:

ln(cos(10.0 rad/s * 5.00 s + delta)) + b*5.00 s / 0.0500 kg / 2 - ln(cos(delta)) = ln(3.00)

We can rearrange this equation to solve for b:

b = (2/5.00 kg) * (ln(3.00) - ln(cos(10.0 rad/s * 5.00 s + delta)) + ln(cos(delta)))

The phase angle delta is unknown, but it cancels out when we take the difference between the two equations for A. Therefore, we can choose any value of delta and still get the correct value of b.

Let's choose delta = 0 for simplicity. Plugging in the values, we get:

b = (2/0.0500 kg) * (ln(3.00) - ln(cos(10.0 rad/s * 5.00 s)) + ln(cos(0)))

b = 0.277 Ns/m

Therefore, the magnitude of the damping coefficient is = 0.277 Ns/m

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A 50.0-g hard-boiled egg moves on the end of a spring with force constant k=25.0N/m. It is released with an amplitude 0.300 m. A damping force Fx=−bv acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.Calculate the magnitude of the damping coefficient b.

URGENT PLEASE HELP SOLVE THANK YOU SO MUCH
centers of two small particles charged particles are separated by a distance of 1.2x10-4 meter. The charges on the particle are 8x10^-19 and 4.8*10^-19, respectively. what is the the magnitude of the electrostatic force between these two particles?

Answers

Answer:

the magnitude of the electrostatic force between these two charged particles is 4.32 x 10^-28 N.

Explanation:

F = k * (q1 * q2) / r^2

where:

F is the electrostatic force

k is the Coulomb's constant (9 x 10^9 Nm^2/C^2)

q1 and q2 are the charges of the particles

r is the distance between the particles

Plugging in the given values:

q1 = 8 x 10^-19 C

q2 = 4.8 x 10^-19 C

r = 1.2 x 10^-4 m

k = 9 x 10^9 Nm^2/C^2

we can calculate the electrostatic force:

F = (9 x 10^9 Nm^2/C^2) * (8 x 10^-19 C) * (4.8 x 10^-19 C) / (1.2 x 10^-4 m)^2

F = 4.32 x 10^-28 N

The electrostatic force between these two particles is therefore calculated to be **1.44x10⁻¹⁴ N**1 using Coulomb's equation of electrostatic force.

COULOMB LAW BE DEFINED?

With a force proportional to the product of the charges and inversely proportional to the square of the distance between them, Coulomb's law states that like charges repel and opposite charges attract. There is a formula that may be used to calculate the force between two point charges. Charles-Augustin de Coulomb created Coulomb's law in 1785

The electrostatic force between two charges is inversely proportional to the square of the distance between them and directly proportional to the product of their magnitudes1.

The centers of the two tiny charged particles in this example are separated by a distance of **1.2x10⁻⁴ meter**, with charges of **8x10⁻¹⁹** and **4.8*10⁻¹⁹**, respectively.

F is the electrostatic force

k is the Coulomb's constant (9 x 10⁹ Nm²C²)

q1 and q2 are the charges of the particles

r is the distance between the particles

Plugging in the given values:

q1 = 8 x 10⁻¹⁹ C

q2 = 4.8 x 10⁻¹⁹C

r = 1.2 x 10⁻⁴ m

k = 9 x 10₉ Nm²/C²

we can calculate the electrostatic force:

F = (9 x 10⁹Nm²/C²) * (8 x 10⁻¹⁹ C) * (4.8 x 10⁻¹⁹ C) / (1.2 x 10⁻⁴ m)²

The electrostatic force between these two particles is therefore calculated to be **1.44x10⁻¹⁴N**1 using Coulomb's equation of electrostatic force.

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By how much would a windmill power increase in [%] if the wind speed increased 3%?

a. 33 b. 9 c. 13 d. 5 e. 21

Answers

The power will increase by about 9.27% (when rounded to two decimal places).

Therefore, the correct answer choice is (b) 9.

How do we calculate?

The power generated by a windmill is proportional to the cube of the wind speed and which means that if the wind speed increases by 3%, the power generated will increase by:

(1.03)^3 −1 = 0.0927

In physics, power is described as the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt.

So the power will increase by about 9.27%

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String is wrapped around an object of mass M = .5 kg and the moment of inertia I = .02 kg*m^2. You pull the string with your hand straight up with some constant force F such that the center of the objects does not move up or down, but the object spins faster and faster. This is like a yo-yo; nothing but the vertical string touches the object. When your hand is a height y = .22 m above the floor, the object has an angular speed of w = 15 radians/s. When your hand has risen to a height of y = .73 m above the floor, what is the angular speed of the object? Your answer must be numeric and not contain the symbol F.

Answers

The angular speed of the object when your hand has risen to a height of y = 0.73 m above the floor is approximately 26.83 radians/s.

What is Mass?

Mass is a fundamental property of matter that quantifies the amount of matter in an object. It is a scalar quantity that describes the resistance of an object to acceleration when subjected to an external force. The mass of an object is a measure of the total amount of material or substance that it contains, and it is usually expressed in kilograms (kg) or other appropriate units.

The final potential energy is given by:

PE_final = m * g * y_final

where y_final is the final height of the object above the floor.

Setting the initial mechanical energy equal to the final mechanical energy and rearranging the equation, we get:

g = 9.8 m/[tex]s^{2}[/tex] (acceleration due to gravity)

(1/2) * 0.02 * [tex](15)^{2}[/tex]^2 + 0.5 * 9.8 * 0.22 = (1/2) * 0.02 * [tex]w_final ^{2}[/tex] + 0.5 * 9.8 * 0.73

w_final = sqrt((2 * ((1/2) * 0.02 * (15)^2 + 0.5 * 9.8 * 0.22 - 0.5 * 9.8 * 0.73)) / 0.02)

Plugging in the given values and solving for w_final, we get:

w_final ≈ 26.83 radians/s (rounded to two decimal places)

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Determine the percentage of kinetic energy lost by a small ball when it makes an elastic head-on collision with stationary bigger ball. The mass of the bigger ball is 12 times bigger than the mass of the small ball.

Answers

The small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.

[tex]\frac{Kf}{Ki} =[/tex] [tex]1- [\frac{1}{2}][ \frac{M}{m}] [\frac{2mv^{2} }{[M+m]^{2} }[/tex]

M = 12m

Kf/Ki = 1 - (1/2)(12m/m)[(2mv^2)/(13m)^2]

Kf/Ki = 165/169

(1 - Kf/Ki) x 100% = (1 - 165/169) x 100% = 2.37%

So, the small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.

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If the wave pictured above oscillates up and down 25 time in 10
seconds, what is the frequency?

Answers

Answer:

Frequency = 25 / 10 = 2.5 Hz

Explanation:

If the wave oscillates up and down 25 times in 10 seconds, then the frequency can be calculated as follows:

Frequency = Number of oscillations / Time

In this case, the number of oscillations is 25, and the time is 10 seconds. Therefore, the frequency is:

Frequency = 25 / 10 = 2.5 Hz

So the frequency of the wave is 2.5 Hertz, which means it completes 2.5 cycles or oscillations in one second.

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