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Evgesh-ka [11]
3 years ago
15

How were lion fish introduced to the areas around Florida

Physics
2 answers:
mr_godi [17]3 years ago
5 0

Answer:

<h3>I think this will answer your question. This is information is not mine and this rightfully belongs to <u>columbia.edu.</u></h3><h3><u /></h3>

This brightly colored fish is native to the Indo-Pacific from Australia north to southern Japan and south to Micronesia.  The lionfish is usually found in coral reefs of tropical waters, hovering in caves or near crevices.  Native regions as well as Savannah, Georgia; Palm Beach and Boca Raton, Florida; Long Island, New York; Bermuda and possibly Charleston. In southern Florida and off the coast of the Carolinas in early to mid 1990s.

<h3><u /></h3>
nekit [7.7K]3 years ago
3 0

Answer:I really don't know sorry

Explanation:

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Pls help mee
NISA [10]

Answer:

the Arrow X shows the direction of amplitude

Explanation:

As the amplitude is the maximum displacement of a wave from the mean position

8 0
2 years ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
What happens to the direction of the line joining when the object slows down ?Explain the observations​
denis-greek [22]

Answer:

it will be curved as in deceleration

Explanation:

5 0
3 years ago
A 2.0 kg bird lands on a 1.0 x 10^1 kg bit of tree bark sitting on a frictionless ice-covered pond. The bird’s initial horizonta
Bond [772]

Here in this type of question we can use momentum conservation

It is because we can see there is no external force on the system

So we can use momentum conservation principle

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

here we know that

m_1 = 2 kg

v_{1i} = 6 m/s

m_2 = 1 * 10^1 kg

v_{2i} = 0

now after bird sits on it then final speed of the both will be same

v_{2f} = v_{1f} = v m/s

2*6 + 1*10^1 * 0 = (2 + 1* 10^1) * v

12 = 12*v

v = 1 m/s

so final speed will be 1 m/s

7 0
2 years ago
A charge partides round a 1 m radius circular particle accelerator at nearly the speed of light. Find : (a) The period (b) The c
Scrat [10]

Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

speed=\dfrac{distance}{time}

Let t is the period of the particle.

t=\dfrac{d}{s}

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

t=\dfrac{d}{c}

t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}

t=2.09\times 10^{-8}\ s

(b) On the circular path, the centripetal acceleration is given by :

a=\dfrac{c^2}{r}

a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}

a=9\times 10^{16}\ m/s^2

Hence, this is the required solution.

8 0
3 years ago
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