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3 years ago

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Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 780 m diameter rotatin

g about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?

1 answer:

ser-zykov [4K]3 years ago

5 0

Answer:1.513 rps

Explanation:

Given

Diameter of cylindrical space $d=780\ m$

When the station rotates it creates centripetal acceleration which is given by

$a_c=\omega ^2r$

Now it must create the effect of gravity so

$g=\omega ^2r$

$\omega =\sqrt{\frac{g}{r}}$

$\omega =0.158\ rad/s$

and $\omega =\frac{2\pi N}{60}$

Thus $N=\frac{4.755}{3.142}$

$N=1.513\ rps$

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b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept

olasank [31]

Answer:

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Explanation:

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3 years ago

Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions

iren [92.7K]

Answer:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

Explanation:

For this case we have the following info given:

Number of Na+ ions $5.7 x10^{11} ions$

Each ion have a charge of +e and the crage of the electron is $1.6 x10^{-19}C$

The time is given $t = 7 ms$ if we convert this into seconds we got:

$t = 7ms * \frac{1s}{1000 ms}= 0.007s$

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

$Q = Ne$

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

$Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C$

And from the general definition of current we know that:

$I =\frac{Q}{t}$

And since we know the total charge Q and the time we can replace:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

The current during the inflow charge in the meter axon for this case is $13.02 \mu A$

3 0

3 years ago

The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.

Otrada [13]

Answer:

a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

I = ∫ F dt = Δp

F t = m $v_{f}$ - m v₀

F = m (v_{f} -v₀o) / t

in the exercise indicates that the speed module is the same, but in the opposite direction

F = m (-2v) / t

if we use Newton's second law

F = m a

we substitute

- 2 mv / t = m a

a = - 2 v / t

let's calculate

a = - 2 4.59 10²/3 10⁻¹³

a = 3.06 10¹⁵ m / s

b) F= m a

F= 4.68 10⁻²⁶ 3.06 10¹⁵

F= 1.43 10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

F_total = n F

F_total = n m a

F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵

F_total = 14.32 10⁻²⁶ N

8 0

3 years ago

During a particular thunderstorm, the electric potential difference between a cloud and the ground is vcloud - vground = 3.50 10

wariber [46]

<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>

8 0

3 years ago

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

5 0

3 years ago