Question #c843a

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Question #c843a

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Answer 1 · 2017-07-10T15:59:38

$v = 25 m s^{- 1}$ $v=25ms^-1$

Explanation:

$m = 1500 k g$ $m=1500kg$
$v_{0} = 15 m s^{- 1}$ $v_(0)=15ms^-1$
$F_{1} = 1000 N$ $F_1=1000N$
$t = 15 s$ $t=15s$

Before a force acting on it, the object is at constant velocity (assuming no other forces is acting on it: $e g .$ $eg.$ friction), thus
$N e t F = 0$ $NetF=0$

After a force acting on it, the object experiences a net force of same magnitude as the applied force.
$F = m a$ $F=ma$
$1000 = 1500 a$ $1000=1500a$
$a = \frac{2}{3} m s^{- 2}$ $a=2/3 ms^-2$

$v = v_{0} + a t$ $v=v_0+at$
$v = 15 + \frac{2}{3} \cdot 15$ $v=15+2/3*15$
$v = 15 + 10$ $v=15+10$
$v = 25 m s^{- 1}$ $v=25ms^-1$

Answer 2 · 2017-07-10T16:10:41

Its velocity will be 25m/s

Explanation:

We can use two known equations to solve this problem.
These are
$F = m \cdot a$ $F=m*a$
$v = u + a \cdot t$ $v=u +a*t$

Lets substitute the values
F=1000N
m=1500 kg
then we have $a = \frac{F}{m} = \frac{1000}{1500}$ $a=F/m=1000/1500$ =2/3 m/ $s^{2}$ $s^2$

Also
u=15 m/s
t=15 s
so that $v = u + a \cdot t = 15 + \frac{2}{3} \cdot 15 = 15 + 10 = 25 s$ $v=u+a*t=15 + 2/3*15=15+10=25s$

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Question #c843a

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