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NARA [144]
3 years ago
13

Astronaut Mark Kelly and his crew traveled to the International Space Station in the space shuttle. The shuttle burned off all r

emaining fuel before entering Earth's atmosphere. When the shuttle returned to Earth it landed at a speed of approximately 354 km/hr or 220 mi/hr, about 20 times the landing speed of a commercial aircraft. What accounts for the shuttle's speed upon landing?
Physics
2 answers:
Snowcat [4.5K]3 years ago
7 0

A) Earth's atmosphere  

B) initial thrust of take-off  

C) angle of approach upon landing  

D) gravitational potential energy

It would be D. Gravitational potential energy As the shuttle prepares to return to Earth it reduces its speed so that gravity pulls it out of its orbit and toward Earth. As it approaches the earth the gravitational potential energy is converted to kinetic energy. The drag of Earth's atmosphere slows the shuttle's descent and causes some kinetic energy to convert to heat energy.

Nitella [24]3 years ago
6 0

A couple of things, if the shuttle burned all of it's fuel before entering Earth's atmosphere then that means that the shuttle was accelerating towards Earth until it ran out of fuel.  At that point, there is little to no air resistance (friction) by the lack of an atmosphere so it keeps accelerating  due to Earth's gravitational force.  The closer the shuttle gets to Earth the stronger the gravitational pull the shuttle experiences.  Note that, once the shuttle reaches Earth's atmosphere it will cause significant amount of friction and thus will cause the shuttle to slow down.

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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
1. why is<br> experimentation so important to science
Alekssandra [29.7K]

Answer: to provide evidence to a theory

Explanation: Experimentation allows for multiple trials to provide evidence to a scientific theory. Without experimentation there would be no data to back up your hypothesis.

5 0
3 years ago
Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0
3 years ago
What does a battery powered heater warms up from ?
Westkost [7]
Battery based heaters use electric resistance heating, which uses a lot of current (electricity) to create the heat.
8 0
3 years ago
What mistake did Carl make?
True [87]

Answer:

He did not multiply the chlorine and oxygen atoms by the coefficient 4

Explanation:

8 0
3 years ago
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