Astronaut 1 has a mass of 75 kg. Astronaut 2 has a mass of 80 kg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in N). Astro 1 visits a planet with gravitational acceleration 12 m/s2. What must be Astro 2 planet's ag to equal Astro 1's weight

Answer:

The direction could change

Answer:

a)   F = 1.70 10⁻⁹N,   F = 1.47 10⁻⁸ N,

b) * the electronegative repulsion, from the repulsion by quantum effects

Explanation:

a) The atraicione force comes from the electric force given by Coulomb's law,

           F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

divalent atoms

In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given

           F = k q² / r²

           F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]

           F = 1.70 10⁻⁹N

monovalent atoms

in this case the load is q = q₀

           F = 2 \ 10^9 \  \frac{ (1.6 \  10^{-19} )^2}{ (0.125 10^{-9} )^2 }

           F = 1.47 10⁻⁸ N

b) repulsive forces come from various sources

* the electronegative repulsion of positive nuclei

* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together

* from the repulsion of electron clouds, by quantum effects

Answer: The mean value = 9.85m/s².

Explanation:

Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]

The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.

Number of measurements =9

Sum of measurements =  88.69

Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]

Hence, the mean value = 9.85m/s².

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

Answer:

4m/s^2

Explanation:

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

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Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.