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Mazyrski [523]
4 years ago
14

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

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 mechanicel energy of the Earth-rocket system?
a. Rocket Kinetic Energy-increasing System Gravitational Potential Energy-increasing System Mechanical Energy-constant
b. Rocket Kinetic Energy-decreasing. System Gravitational Potential Energy-increasing Sysem Mechanical Energy-constant
c. Rocket Kinetic Energy-Increasing System Gravitational Potential Energy-decreasing Sysem Mechanical Energy-decreasing
d. Rocket Kineic Energy-Increasing System Gravitational Potential Energy nreasing Syste Mechanical Energy-increasing
Physics
1 answer:
poizon [28]4 years ago
4 0

Answer:

c) True.    Rocket Kinetic Energy-Increasing System Gravitational Potential Energy-decreasing Sysem Mechanical Energy-decreasing

Explanation:

Let's write the three equations.

The kinetic energy is

                      K = ½ m v2

As the speed of the rocket increases the kinetic energy must be increased

The potential energy of the rocket system - Earth

                    U = -G m M / r

As the rocket moves away from the earth the power energy must decrease

Mechanical energy

                 Em = K + U

In this case, if we neglect the energy of the exhaust gases, the mechanical energy remains constant.

if we take into account the energy of the gases, the mechanic energy must decrease due to the loss of the energy that takes them

Let's examine the answers

a) False. The power power l decreases

b) False. The kinetic energy increases

c) True. Meets all of the above

d) False The mechanical energy of the system is constant

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How fast will a ball be traveling if it falls for 5 seconds​
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How to calculate energy needed for a change of state ​
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7 0
3 years ago
Read 2 more answers
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
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