If a rocket with a mass of 3900 tons vertically accelerates at a rate of $\frac{7}{8} \frac{m}{s^{2}}$ $7/8 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 5 seconds?

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If a rocket with a mass of 3900 tons vertically accelerates at a rate of $\frac{7}{8} \frac{m}{s^{2}}$ $7/8 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 5 seconds?

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Answer 1 · 2018-07-08T13:34:21

Approximately $182.14$ $182.14$ megawatts.

Explanation:

We first find the net force acting on the rocket using Newton's second law of motion, which states that,

$F_{net} = m a$ $F_"net"=ma$

where:

$F_{net}$ $F_"net"$ is the net force in newtons

$m$ $m$ is the mass in kilograms

$a$ $a$ is the acceleration in meters per second squared

Here, the net force is the force applied to the rocket by acceleration plus the rocket's weight.

$:.F_"net"=ma+mg$

$=m(a+g)$

Substituting our values, we get:

$F_"net"=3900000 \ "kg"(0.875 \ "m/s"^2+9.8 \ "m/s"^2)$

$=41632500 \ "N"$

Now, we find the velocity of the rocket, given by the equation:

$v=u+at$

where:

$v$ is the final velocity

$u$ is the initial velocity

$a$ is the acceleration

$t$ is the time taken

Assuming $u=0$ , we find the rocket's speed after five seconds:

$v=0+0.875 \ "m/s"^2*5 \ "s"$

$=4.375 \ "m/s"$

Power is given by the equation:

$P=F*v$

where:

$P$ is the power in watts

$F$ is the force in newtons

$v$ is the velocity in meters per second

So, we get:

$P=41632500 \ "N"*4.375 \ "m/s"$

$=182142188 \ "W"$

$~~182.14 \ "MW"$

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If a rocket with a mass of 3900 tons vertically accelerates at a rate of $\frac{7}{8} \frac{m}{s^{2}}$ $7/8 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 5 seconds?

1 Answer

Explanation:

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If a rocket with a mass of 3900 tons vertically accelerates at a rate of $\frac{7}{8} \frac{m}{s^{2}}$ $7/8 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 5 seconds?