A 1500 kg weather rocket accelerates upward at 10 m/s2 . It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m above the ground. For help with math skills, you may want to review: Solving Algebraic Equations

V2=-20.0155m/s

Step-by-step explanation:

To answer this question we will use the principle of conservation of momentum which is

MtVt=M1V1+M2V2

1) let's find the initial velocity.

We will use,

v=a*t

v=10m/s2*2s = 20 m/s

d=vot+1/2*a*t2

d=0+1/2*10*4

d=20m

2) For the speed of the lighter piece we use,

v22=v12+2ad

0=v12+2*-9.81*(530-20)

v12=10006.2

v1=100.031m/s

3) Computing this in our conservation of momentum equation it will be-

MtVt=M1V1+M2V2

1500kg*20m/s=(1500*1/3)kg*100.031m/s + (1500*2/3)+V2

30000=50015.5+1000V2

Therefore V2=-20.0155m/s.

step-by-step explanation: noo

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form