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Sergeeva-Olga [200]
3 years ago
12

Find the following answer based on the image.

Physics
1 answer:
saul85 [17]3 years ago
5 0

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷)                      [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

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An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over
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Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air =  1.3 kg/m³

weight of plane

From Bernoulli's principle

P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2

where 1 and 2 are two different locations at the same geo potential level  

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)

\Delta P =1227.2\ N/m^2

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

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A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t
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A) 2.4\cdot 10^{-16}kg

The radius of the oil droplet is half of its diameter:

r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m

Assuming the droplet is spherical, its volume is given by

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3

The density of the droplet is

\rho=885 kg/m^3

Therefore, the mass of the droplet is equal to the product between volume and density:

m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg

B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

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