Answer: 0.790 g/cm3
Explanation:
The density of acetone is 790 Kg/m3.
To convert from Kg to g we multiply by 1000 (1 Kg = 1000 g)
To convert from m3 to cm3 we multiply by 10∧6
So, The density of acetone in (g/cm3) = (790 x 1000) / (10∧6) = 0.79 g/cm3
The answer is D. Products are formed from reactants by the breaking and forming of new bonds.
Answer:
For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive
Explanation:
Solution
Given that:
From the question stated, Anna drew a diagram to compare forces that are strong and weak.
Now,
We are to find which labels are grouped in areas marked as X, Y, Z respectively.
Thus,
For X, Y, Z it is marked as:
X: Always attractive or attractive only
Y: Very small range
Z: Repulsive and attractive
Answer:
Explanation:
extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .
kx = mg
k is spring constant , x is extension , m is mass
k x 8.6 x 10⁻² = 7.52 x 9.8
k = 856.93 N/m
= 857 x 10⁻³ KN /m
b ) Both side is pulled by force of 188 N .
Tension in spring = 188N
kx = T
856.93 x = 188
x = .219.38 m
= 21.938 cm
= 21.9 cm .
length of spring = 31.8 + 21.9
= 53.7 cm .
Answer:
v = 5.15 m/s
Explanation:
At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N
As the cable tension is less than this value, the car must be accelerating downward.
7730 = 984(9.81 - a)
a = 1.95 m/s²
kinematic equations s = ut + ½at² and v = u + at
-5.00 = u(4.00) + ½(-1.95)4.00²
u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s
v = 2.65 + (-1.95)(4.00)
v = -5.15 m/s
