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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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Which is not one of the three forms of energy that travels to earth
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2 years ago
A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass
makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s

Velocity of cart with child on top is 3.38 m/s

7 0
3 years ago
car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down ove
Tomtit [17]

Answer:

78.4 KN/m

Explanation:

Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'=   1 cm  => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e  F=kx

where,

F = the force needed

x= distance the spring is stretched or compressed beyond its natural length

k= constant of proportionality called the spring constant.

F=kx ---> (since f=mg)

mg=kx

k=(mg)/x

k=(80 x 9.8)/ 0.01

k=78.4x10^3

k=78.4 KN/m

3 0
3 years ago
Read 2 more answers
Could anyone help me out ?
Delicious77 [7]

density = mass/volume = 100kg/10ml = 10kg/ml

voluime = mass/density = 50g/2 g/ml = 25 ml

mass = density x volume = 2x55 = 110 kg

4 0
2 years ago
Read 2 more answers
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