Acceleration of car is 4m/s²
Acceleration of truck is 2m/s²
Explanation:
Given parameters:
Force on the car and truck = 12000N
Mass of car = 3000kg
Mass of truck = 6000kg
Unknown:
Acceleration of car = ?
Acceleration of truck = ?
Solution:
Force is product of mass and acceleration. It is a push or a pull exerted on a body.
Force = mass x acceleration
Make acceleration the subject of the formula;
Acceleration = 
Acceleration of car = 
= 4m/s²
Acceleration of truck = 
= 2m/s²
learn more:
Acceleration brainly.com/question/3820012
Newton's law brainly.com/question/11411375
#learnwithBrainly
Answer:
materials: solid dissolved in a liquid
independent variable: temperature
Explanation:
I got it right
Answer:
a) W₁ = 54000 Lb-ft
b) W₂ = 77760 Lb-ft
c) W₃ = 24000 Lb-ft
W₄ = 40560 Lb-ft
Step by step
W= ∫₁² ydF 1 and 2 are the levels of liquid
Where dF is the differential of weight of a thin layer
y is the height of the differential layer and
ρ*V = F
Then
dF = ρ* A*dy*g
ρ*g = 60 lb/ft³
A= Area of the base then
Area of the base is:
A(b) = 4*2 = 8 ft²
Now we have the liquid weighs 60 lb/ft³
Then the work is:
a)
W₁ = ∫₀¹⁵ 8*60*y*dy ⇒ W₁ =480* ∫₀¹⁵ y*dy
W₁ =480* y² /2 |₀¹⁵ ⇒ 480/2 [ (15)² - 0 ]
W₁ = 240*225
W₁ = 54000 Lb-ft
b) The same expression, but in this case we have to pump 3 feet higher, then:
W₂ = ∫₀¹⁸ 480*y*dy ⇒ 480*∫₀¹⁸ydy ⇒ 480* y²/2 |₀¹⁸
W₂ = 480/2 * (18)²
W₂ = 240*324
W₂ = 77760 Lb-ft
c) To pump two-thirds f the liquid we have
2/3* 15 = 10
W₃ = 480*∫₀¹⁰ y*dy ⇒ W₃ = 480* y²/2 |₀¹⁰
W₃ = 240*(10)²
W₃ = 24000 Lb-ft
d)
W₄ =480*∫₀¹³ y*dy
W₄ =480* y²/2 |₀¹³
W₄ = 240*(13)²
W₄ = 240*169
W₄ = 40560 Lb-ft
They end up cancelling each other out in a process called destructive interference
