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3 years ago

What is the WWF doing to help protect turtle eggs? (Site 1)

Physics

1 answer:

ivolga24 [154]3 years ago

4 0

<h3><u>WWF in protection of Turtle eggs:</u></h3>

World Wildlife Fund (WWF) is an NGO that primly aims wildlife preservation. To protect marine turtles and avoid its poaching, WWF works through TRAFFIC, the Wildlife trade monitoring network.

Turtle eggs are highly demanded for human consumption, either raw or as expensive dishes in star rated restaurants. WWF along with TRAFFIC promotes ecotourism by raising awareness to stop preferring turtle products. It establishes and strengthens MPA's (Marine Protected areas) areas where turtle nesting is primarily observed and thus turtles are free and safe to nest, feed and migrate.

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Zepler [3.9K]

Answer:

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Explanation:

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3 years ago

A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters

emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

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This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

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3 years ago

After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri

Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, $\Delta x=0.03\ m$

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

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Substituting all the values in above equation,

$\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2$

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

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2 years ago

Seesaw unbalanced force explain

wolverine [178]

Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put

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Otrada [13]

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