<span>Answer:
F(x) = ax^2 - bx
or
F(x) = ax² - bx
F(x) = 30x² - 6x
â«F(x)dx = â«(30x² - 6x)dx
as this is evaluated from zero to x
W = 10x³ - 3x² <===ANS
W = 10(0.42³) - 3(0.42²) - [10(0³) - 3(0²)]
W = 0.212 J <===ANS
W = 10(0.72³) - 3(0.72²) - [10(0.42³) - 3(0.42²)]
W = 1.966 J <===ANS</span>
Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
I believe Intangibility is the answer! :P I hope this helps!
Explanation:
Formula for calculating the area of a rectangle A = Length *width
For statement A;
Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
Area of the rectangle = 2.536mm * 1.4mm
Area of the rectangle = 3.5504mm²
The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.
Area of the rectangle = 3.6mm² (to 2sf)
For statement B;
Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.
Area of the rectangle = 2.536mm * 1.41mm
Area of the rectangle = 3.57576mm²
Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.
Area of the rectangle = 3.58mm² (to 3sf)
Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.
Answer:
Imp = 25 [kg*m/s]
v₂= 20 [m/s]
Explanation:
In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.
1)
where:
m₁ = mass of the object = 5 [kg]
v₁ = initial velocity = 0 (initially at rest)
F = force = 5 [N]
t = time = 5 [s]
v₂ = velocity after the momentum [m/s]
![(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]](https://tex.z-dn.net/?f=%285%2A0%29%20%2B%285%2A5%29%20%3D%20%28m_%7B1%7D%2Av_%7B2%7D%29%20%3D%20Imp%5C%5CImp%20%3D%2025%20%5Bkg%2Am%2Fs%5D)
2)
![(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29%2B%28F%2At%29%3D%28m_%7B1%7D%2Av_%7B2%7D%29%5C%5C%280.075%2A0%29%2B%2830%2A0.05%29%3D%280.075%2Av_%7B2%7D%29%5C%5Cv_%7B2%7D%3D20%20%5Bm%2Fs%5D)
