If a rocket with a mass of $700 k g$ $700 kg$ vertically accelerates at a rate of $\frac{3}{2} \frac{m}{s^{2}}$ $3/2 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

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If a rocket with a mass of $700 k g$ $700 kg$ vertically accelerates at a rate of $\frac{3}{2} \frac{m}{s^{2}}$ $3/2 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

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Answer 1 · 2017-01-29T14:47:35

P=6.3 KW

Explanation:

see

MASS=700 kg
Acceleration= $\frac{3}{2} \frac{m}{s^{2}}$ $3/2m/s^2$

THEREFORE force exerted is $m \cdot a = 700 \cdot \frac{3}{2} = 1050 N$ $m*a=700*3/2=1050N$

Now

$a = \frac{v - u}{t}$ $a=(v-u)/t$ and we have U ,i.e initial velocity =0 as it started from rest and ofcourse every rockets before launching is at rest,lol.

so by this we can get $v$ $v$ and $t$ $t$ we are given 4 seconds

$\frac{3}{2} = \frac{v}{4}$ $3/2=v/4$
$v = 6 \frac{m}{s}$ $v=6 m/s$

therefore

$E = F \cdot D i s p l a c e m e n t$ $E=F*Displacement$
dividing both sides of the equation by $t$ $t$

$\frac{E}{t} = F \cdot \frac{d i s p l a c e m e n t}{t}$ $E/t=F*(displacement)/t$

And by definition power is energy per unit time and velocity is displacement per unit time.so applying these

$p = F \cdot v$ $p=F*v$
$P = 1050 \cdot 6$ $P=1050*6$ Watt

$P = 6300 W$ $P=6300 W$
$P = 6.3 K W$ $P= 6.3 KW$

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If a rocket with a mass of $700 k g$ $700 kg$ vertically accelerates at a rate of $\frac{3}{2} \frac{m}{s^{2}}$ $3/2 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

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Explanation:

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If a rocket with a mass of $700 k g$ $700 kg$ vertically accelerates at a rate of $\frac{3}{2} \frac{m}{s^{2}}$ $3/2 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 4 seconds?