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ololo11 [35]
2 years ago
13

If the average speed of an orbiting space shuttle is 27 800 km/h, determine the time required for it to circle Earth. Assume tha

t the shuttle is orbiting about 320.0 km above Earthâs surface, and that Earthâs radius is 6380 km.
Physics
1 answer:
balu736 [363]2 years ago
7 0

Answer:

 t = 1.51 hours

Explanation:

given,

Speed of space shuttle. v = 27800 Km/h

Radius of earth, R = 6380 Km

height of shuttle above earth, h = 320 Km

Total radius of the shuttle orbit

r' = R + h

r' = 6380 + 320

r' = 6700 Km

distance, d = 2 π r

   d = 2 π x 6700

time = \dfrac{distance}{speed}

time = \dfrac{2\pi\times 6700}{27800}

 t = 1.51 hours

Time require by the shuttle to circle the earth is equal to 1.51 hr.

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D. that looks like a rubber band.
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2 years ago
1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's
Anit [1.1K]

Answers:


1. Firstly, we have to define that Pressure P is Force applied F per unit area A. It is mathematically expressed as follows:


P=\frac{F}{A}   (1)


The unit of P is Pascal (Pa) which is equivalent to \frac{kg}{ms^{2}} and also equivalent to \frac{N}{m^{2} }


There is also another expression of the Pressure in which it is dependent on the density d  of the liquid, the height h of the container and the gravity force g:


P=d*h*g     (2)


In this problem the liquid is water, and its known density is approximately:


d=1000kg/m^{3}


So, we have to substitute the values in equation (2) to obtain the pressure <u>(Being careful with the units)</u>:


P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}


P=98000Pa


Then, we have to substitute this value in equation (1) and clear F:


F=P*A


Finally:

F=196000N



2. For this problem, we will use equation (1) to find the Pressure. We already know the area A and the force exerted by water in the container F:


P=\frac{F}{A}=\frac{900N}{3m^{2}}


P=300Pa


3. In this case, equation (2) is the perfect way to find the hydrostatic pressure at any point at the bottom of the tank <u>(be careful with the units):</u>


P=d*h*g      

P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}      


P=73500Pa


4. In this case, it's important to know that in fluids (in this case the water) the higher the fluid is, the lower the pressure. Then, if P_{1} and P_{2} are the respective pressures at the heights h_{1} and h_{2}, and knowing that the water density and the gravity force in this case are constants, we can use the following expression to solve this problem:


P_{2}- P_{1} =d*g(h_{2}- h_{1})   (3)


Where:


P_{1}=1.5 kPa at h_{1}=2m


Note that 1kPa=1*1000 Pa


And P_{2}=? is unknown at a given height h_{2}=6m


Then, we have to substitute the values in equation (3) to find P_{2}:


P_{2}-1500Pa=1000\frac{kg}{m^{3}}*9.8\frac{m}{s^{2}} (6m-2m)    


Finally: P_{2} =40700Pa    


5. In this case we have the area A=0.75m^{2} and the mass of the piston m=200kg, and we need to know the pressure P.


We will use equation (1):  

P=\frac{F}{A}


But, <u>do you remember that above we stated that pressure is the force applied over an area?</u>

Well, in this case we will use the following equation, in which the gravity force and the mass of a body are involved, to find F:


F=m*g=200kg*9.8\frac{m}{s^{2}}


Then:


F=1960N


Now we can finally calculate P:


P=\frac{1960N}{0.75m^{2}}


P=2613.33Pa



5 0
2 years ago
Read 2 more answers
A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.
Natali5045456 [20]

Answer:

option (c)

Explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

6 0
2 years ago
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A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

4 0
2 years ago
An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen
Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration a=1.1 m/s^2

Force exerted by Rope is equal to Tension in the rope

F_{net}=T=ma_{net}

T=\frac{1040}{g}\times 1.1

T=116.84 N

8 0
2 years ago
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