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Sunny_sXe [5.5K]

3 years ago

9

a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being

zero is started underwater at -35 ft jump up and out of the water the 27 ft and return to the water to swim at -18 feet what is the total vertical distance the dolphin traveled?

2 answers:

Yuki888 [10]3 years ago

5 0

Answer:

107 feet is the total vertical distance the dolphin traveled.

Explanation:

Negative sign in front of the distance value is just to define the distance covered by dolphin underwater that is beneath the surface of the water

Before jump

Vertical distance covered by dolphin below the surface of water before coming out= 35 feet

Vertical distance covered by dolphin after coming out of the surface of water = 27 feet

After jump

Vertical distance covered by dolphin above the surface of water after jumping = 27 feet

Vertical distance covered by dolphin below the surface of water after jumping = 18 feet

The total vertical distance the dolphin traveled:

35 feet + 27 feet + 27 feet + 18 feet = 107 feet

marin [14]3 years ago

3 0

From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

35 ft + 27 ft + 27 ft + 18 ft = 107 ft

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What is the science behind the making of pop-up books?

IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0

3 years ago

What traction of the radioisotope<br>remains in the body after one day?

r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is $(\frac{1}{2})^{\frac{1}{\tau}}$ , with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

$\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}$

where the half-life $\tau$ must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0

3 years ago

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

6 0

3 years ago

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Gekata [30.6K]

Answer:

A.

Explanation:

an atom that has most of its mass is electrons.

6 0

3 years ago

Which technological advance allows scientists to handle these objects enough to feel their properties while still protecting the

muminat

The answer depends heavily on what 'objects' you're talking about.

5 0

3 years ago

Read 2 more answers