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kogti [31]
3 years ago
10

A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass o

f fuel ejected is small compared to the mass of the rocket. For this distance, which of the following indicates the changes, if any, in the kinetic energy of the rocket, the gravitational potential energy of the Earth-rocket system, and the mechanical energy of the Earth-rocket system?
1. KE Increases, PE Increases, ME Increases
2. KE Increases, PE Increases, ME Constant
3. KE Increases, PE Decreases, ME Decreases
4. KE Decreases, PE Increases, ME Constant
Physics
2 answers:
mamaluj [8]3 years ago
4 0

Answer:

The correct answer is

1. KE Increases, PE Increases, ME Increases

Explanation:

Here, it is essential to define the following concepts.

Kinetic energy is the energy due to motion of an object. The definition of kinetic energy is the energy to accelerate a body from rest to a given velocity. The body retains the kinetic energy as it maintains the attained velocity

KE= 1/2mv^2

From the above definition KE is increasing

Potential energy is the energy stored within a body due to its location in a position or location of higher potential

As an object moves up to higher altitude in a direction opposite to the earth's attractive force, its gravitational potential increases

PE=mgh

From the above definition PE is increasing

Mechanical energy is the sum of potential and kinetic energies that can be used to do work. It is the sum of the potential and kinetic energies stored in the body

ME = PE+KE

Since both PE and KE are increasing then ME is also increasing

TEA [102]3 years ago
4 0

Answer:

1. KE Increases, PE Increases, ME Increases.

Explanation:

As the rocket is continuously firing its engines as it accelerates away from Earth, it means that the mechanical energy (kinetic energy + potential energy) of the Earth-rocket system increase.

As mass decreases, kinetic energy decreases if all other variables are constant.

Also as the rocket is moving away from the Earth its potential energy due to gravity known as gravitational potential energy also increases.

 

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t
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Answer:

final velocity =  0.08585m/s

Explanation:

We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.

The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after

total linear momentum before = linear momentum of first train car + linear momentum of second train car

We know that linear momentum = mv

where,

m = mass

v = velocity

thus,

total linear momentum before = m₁v₁ + m₂v₂

m₁ = mass of first train car = 135,000kg

v₁ = velocity of first train car = 0.305m/s

m₂ = mass of first second car =  100,000kg

v₂ = velocity of second train car =  −0.210m/s

Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.

total linear momentum before = m₁v₁ + m₂v₂

                                                  = 135,000x0.305 + 100,000x(−0.210)

                                                  = 135,000x0.305 - 100,000x0.210

                                                  = 20,175 kgm/s

Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).

Therefore, total linear momentum after = mv

m = m₁ + m₂ = 135,000 + 100,000 = 235,000

total linear momentum before = total linear momentum after

235,000v = 20,175

v =  \frac{20,175}{235,000}

  = 0.08585m/s

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3 years ago
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Explanation:

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3 years ago
A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.
astra-53 [7]

Answer:

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be  -0.5026m/sec^2

Explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

S=ut+\frac{1}{2}at^2

40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2

7.45u+27.751a=40-----eqn 1

According to first equation of motion

v = u + at

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Solving equation 1 and 2

a = -0.5026m/sec^2

And u = 7.244 m /sec

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be

-0.5026m/sec^2

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Ahat [919]

Answer:

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P \ V = \ n \ R \ T

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We can use the atmospheric pressure as 1 atm, and the body temperature as 36.5 °C, in Kelvin this is:

T_{body} = 36.5 \° C = (36.5 + 273.15) K = 309.65 \ K

The ideal gas constant is:

R = 0.082057 \frac{L \ atm}{ K \ mol}

taking all this in consideration, the number of moles will be:

n = \frac{P \ V}{  R \ T }

n = \frac{1 \ atm * 3.8 \ L  }{ 0.082057 \frac{L \ atm}{ K \ mol} *  309.65 \ K } [/tex]

n = 0.14955 \ mol

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