If a rocket with a mass of #2500 tons # vertically accelerates at a rate of # 5/2 m/s^2#, how much power will the rocket have to exert to maintain its acceleration at 15 seconds?

Answer 1

The power is #=1153.1MW#

The mass of the rocket is #m=2500000kg#
The acceleration of the rocket is #a=5/2ms^-2#
The acceleration due to gravity is #=9.8ms^-2#

According to Newton's second Law, the net force on the rocket is

#F=m(a+g)=2500000*(9.8+2.5)=30750000N#
Assuming that the initial velocity of the rocket is #u=0ms^-1#
The velocity after #t=15s# is
#v=u+at=0+2.5*15=37.5ms^-1#

The power is

#P=Fv=30750000*37.5=1153125kW=1153.1MW#
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Answer 2

Power exerted by the rocket = Force × Velocity Force exerted by the rocket = Mass × Acceleration Force = (2500 tons) × (5/2 m/s^2) Velocity = Acceleration × Time = (5/2 m/s^2) × (15 s)

Calculate Force, then substitute into the power formula.My apologies for the oversight. To clarify, the correct formula for power (P) is:

[ P = \text{Force} \times \text{Velocity} ]

Given that force (F) is calculated as the product of mass (m) and acceleration (a):

[ F = m \times a ]

And velocity (v) is determined as the product of acceleration (a) and time (t):

[ v = a \times t ]

Now, substitute these values into the power formula:

[ P = (m \times a) \times (a \times t) ]

Plug in the provided values:

[ P = (2500 , \text{tons} \times \frac{5}{2} , \text{m/s}^2) \times \left(\frac{5}{2} , \text{m/s}^2 \times 15 , \text{s}\right) ]

Calculate to find the power exerted by the rocket.

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Answer 3

To calculate the power exerted by the rocket, we can use the formula:

[ P = F \cdot v ]

Where:

  • ( P ) is the power,
  • ( F ) is the force exerted by the rocket,
  • ( v ) is the velocity of the rocket.

The force exerted by the rocket can be calculated using Newton's second law:

[ F = m \cdot a ]

Where:

  • ( m ) is the mass of the rocket,
  • ( a ) is the acceleration.

Given that the mass of the rocket is 2500 tons and the acceleration is ( \frac{5}{2} ) m/s², we can calculate the force:

[ F = 2500 , \text{tons} \times \frac{5}{2} , \text{m/s}^2 ]

Then, we need to convert the mass from tons to kilograms, where 1 ton = 1000 kg:

[ 2500 , \text{tons} = 2500 \times 1000 , \text{kg} ]

Now, we can calculate the force:

[ F = 2500 \times 1000 , \text{kg} \times \frac{5}{2} , \text{m/s}^2 ]

Next, we need to calculate the velocity of the rocket. Since the rocket is accelerating vertically, we'll use the formula:

[ v = u + at ]

Where:

  • ( u ) is the initial velocity (which is typically 0 for objects starting from rest),
  • ( a ) is the acceleration,
  • ( t ) is the time.

Given that the rocket starts from rest, the initial velocity ( u = 0 ). So, we can calculate the final velocity using:

[ v = at ]

Substituting the values, ( a = \frac{5}{2} , \text{m/s}^2 ) and ( t = 15 ) seconds:

[ v = \frac{5}{2} , \text{m/s}^2 \times 15 , \text{seconds} ]

Now, we can calculate the power:

[ P = F \times v ]

Substitute the calculated values for force and velocity into the equation to find the power. Make sure to convert tons to kilograms before performing the calculations.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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