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AURORKA [14]
3 years ago
8

On the surface of planet y, which has a mass of 4.83x1024 kg, a 30.0 kg object weighs 50.0 n. what is the radius of the planet?

Physics
1 answer:
Aleksandr [31]3 years ago
5 0
We can solve the problem in two steps:

1) From the weight W=50.0 N of the object, we can find the value of the gravitational acceleration g of the planet. In fact, the weight is equal to
W=mg
where m=30 kg is the mass of the object. From this, we find g:
g= \frac{W}{m}= \frac{50.0 N}{30 kg}=1.67 m/s^2

2) The gravitational acceleration of a planet with mass M and radius r is given by
g= \frac{GM}{r^2}
where G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational  constant. In our problem, the mass of the planet is 
M=4.83 \cdot 10^{24} kg, and we found g in step 1), g=1.67 m/s^2, so we have everything to solve and find the value of the radius r:
r= \sqrt{ \frac{GM}{g} }= \sqrt{ \frac{(6.67\cdot 10^{-11})(4.83 \cdot 10^{24})}{1.67} }=1.39\cdot 10^7 m
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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
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a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

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k = 4.3 is the dielectric constant

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d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

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Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

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brainly.com/question/10427437

brainly.com/question/8892837

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