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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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On earth which force is 10 to slow an object down
ololo11 [35]

Answer:

gravity

Explanation:

6 0
3 years ago
5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
A container is filled with water to a depth of 26.2 cm. On top of the water floats a 16.1 cm thick layer of oil with a density of
Solnce55 [7]

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

                                                      = 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

4 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
3 years ago
Solar system help please!
Kobotan [32]
When one body(sun) exerts a force on a second body(planet), the second body simultaneously exerts a force equal in magnitude and opposite in direction of the first body. Which makes the planet orbit in path C.

Hope this helps!!
5 0
3 years ago
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