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Verizon [17]
4 years ago
15

A space station with a radius of 120 m rotates once every 70 s to create artificial gravity. If the astronaut has an earth weigh

t of 160 lbs, then what is its artificial weight?
Physics
1 answer:
Alina [70]4 years ago
5 0

Answer:

Artificial weight = 70.27 N = 15.80 lbs

Explanation:

The earth weight of the astronaut = 160 lbs = 711.72 N

The weight on earth = m × g(earth)

g(earth) = 9.8 m/s²

711.72 = m × 9.8

m = (711.72/9.8)

m = 72.62 kg

But at the space station, the space station rotates once every 70 s to create an artificial radial acceleration that creates a radial gravity pulling the objects on the space station towards the centre of that space station.

radial acceleration = α = (v²/r)

v = rw,

α = (rw)²/r

α = rw²

r = radius of rotation = 120 m

w = angular velocity = (2π/70) (it completes 1 rotation, 2π radians, in 70 s)

w = 0.0898 rad/s

α = 120 × (0.0898²)

α = 0.968 m/s²

Artificial weight = (mass of astronaut) × (Radial acceleration) = 72.62 × 0.968

Artificial weight = 70.27 N = 15.80 lbs

Hope this Helps!!!

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You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
Crazy boy [7]

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

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5 0
3 years ago
Why do we need gametes?
andrew11 [14]

Answer:

Gametes are the cells used during sexual reproduction to produce a new individual organism.

Explanation:

The male gamete or sperm, is a smaller, mobile cell that meets up with the much larger and less mobile female gamete, egg or ova.

6 0
3 years ago
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dem82 [27]

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umka21 [38]

Answer:

Explanation:

s = s₀ + v₀t + ½at²

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s = ½10(15²) = 2250 m  if air resistance is ignored

7 0
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