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3 years ago

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The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th

e largest around 19 mm. Mosquitoes typically have a mass between 3.0 and 5.0 mg. All mosquitoes have slender bodies with three segments: a head, a thorax and an abdomen. What is the length of the largest known mosquito in inches?

1 answer:

FrozenT [24]3 years ago

4 0

0.74in

Explanation:

Given parameters:

length range of adult mosquito = 3.0 - 6.0mm

Smallest mosquito = 2.5mm

Largest mosquito = 19mm

Average mass range = 3.0 - 5.0mg

Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

The problem here is converting from mm to inches;

The length of the longest mosquito which is the largest is 19mm

19mm to inches;

1mm = 0.039inches

19mm = 19mm x $\frac{0.039in}{1mm}$ = 0.74in

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Scale brainly.com/question/570757

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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

3 0

3 years ago

You are looking at yourself in a plane mirror, a distance of 3 meters from the mirror. your brain interprets what you are seeing

AleksandrR [38]

You are looking at yourself in a plane mirror, a distance of 3 meters from the mirror. your brain interprets what you are seeing in the mirror as being a person standing 6 meters from you.

<h3>Calculation</h3>

The plane mirror shows an exact replica of the real world. that means the distance of you from the mirror is the same distance as your reflection form the mirror at the opposite side of the mirror.

Thus, distance of image from the plane mirror is same as the distance of object (person) from the plane mirror but the image is formed behind the mirror.

Thereby we have v=u=3 m

Thus, distance between image and the person

is d = v + u = 3 + 3 = 6 m

Thus, the person is 6 meters away from the image.

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8 0

2 years ago

3. If you roll a ball up a hill, it undergoes positive<br>acceleration. True or false<br><br>

Jlenok [28]

Answer:

true

Explanation:

4 0

3 years ago

A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the t

Virty [35]

Answer:

B) d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

Em₀ = U = m g h₁

Final point. Lower (slide bottom)

Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

mgh₁ = ½ m v² + mgh₂

v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

t = √ 2h₂ / g

We substitute in the other equation

d = √ (2g (h₁-h₂)) √ 2h₂ / g

d = √ (4 h₂ (h₁-h₂))

H = h₁ + h₂

h₁ = H -h₂

d = √ ( 4 h₂ ( H - 2h₂))

3 0

3 years ago

Read 2 more answers

which element are you most likely to find as a free element rather than a compound lead or calcium, explain?

TEA [102]

Pb, if compared to Ca it is less reactive and it is a transition metal, and is highly stable alone.

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3 years ago