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posledela
3 years ago
9

Let's see about this one.

Physics
2 answers:
Mashutka [201]3 years ago
8 0

Answer: The answer would be A, D and possible E..

Explanation:

The person will continue breathing of course but, shallow breaths so its not c.

and more waste will travel around the cell because oxygen is use in the cell and the cell distributes waste.. Without air, the cells die and yeahh.. im not very good at explaining sorry^^ but its not b either.

Pachacha [2.7K]3 years ago
6 0

×#(sns,sikam2kqi822hdbd3issjwjxdiwi&×&$>=&÷3838'

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Which depends on your location weight or mass
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I believe it would be weight. mass never changes.
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2 years ago
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A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2
Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

a=\dfrac{v-u}{t}

a=\dfrac{10\ m/s}{1.28\ s}

a=7.81\ m/s^2

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, a=\dfrac{36}{0.000355}

a=101408.45\ km/h^2

Hence, this is the required solution.

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3 years ago
Pluto has been reassigned and is now a dwarf planet. Why did scientists think this reassignment was necessary? If you were a sci
kap26 [50]

Answer:

Explanation:

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Answer:

B

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6 0
3 years ago
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

             θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

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6 0
3 years ago
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