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vitfil [10]
3 years ago
8

Identify the part labeled 1, 2, 3, 4, and 5 of the central nervous system.

Physics
1 answer:
Sindrei [870]3 years ago
7 0
How are we supposed to help answer this if we can't see what parts are labeled? 

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When electrons are removed from the outermost shell of a calcium atom, the atom becomesA. an anion that has a larger radius than
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Answer:

D. a cation that has a smaller radius than the atom.

Explanation:

When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.

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Which data set has the largest range?
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Line the numbers from smallest to largest the subtract the smallest from the largest numbers.

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The product side of a chemical reaction is shown. → 7Ti2(SO4)3
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The answer is the fourth choice because there are 7 represents in a coefficient.

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You are given a parallel plate capacitor that has plates of area 29 cm2 which are separated by 0.0100 mm of nylon (dielectric co
GalinKa [24]

Answer:

20.60 kV

Explanation:

Capacitance of parallel plates without dielectric between them is:

C=\frac{\varepsilon_{0}A}{d}

with d the distance between the plates, A the area of the plates and ε₀ the constant 8.85419\times10^{-12}\frac{C^{2}}{Nm^{2}}, so :

C_0=\frac{(8.85419\times10^{-12})(0.0029)}{0.0100\times10^{-3}}=2.57\times10^{-9} F

But the dielectric constant is defined as:

k=\frac{C}{C_{0}}

With C the effective capacitance (with the dielectric) and Co the original capacitance (without the dielectric). So, the new capacitance is:

C=kC_0

But capacitance is related with voltage by:

C=\frac{Q}{V}

with Q the charge and V the voltage, using the new capacitance and solving for V:

kC_0=\frac{Q}{V}

V=\frac{Q}{kC_0}=\frac{0.18\times10{-3}}{(3.4)(2.57\times10^{-9})}=20599.68V=20.60 kV

4 0
3 years ago
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You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

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6.32 m/s 18.43° northeast

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We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
3 years ago
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