Answer: c
Explanation:
The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.
a. 2×2 atoms of Al ≠ 3×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!
b. 3×2 atoms of Al ≠ 3×1 atoms of Al
3×3 atoms of O ≠ 2×2 atoms of O
c. 2×2 atoms of Al = 4×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH ARE EQUAL, CORRECT ANSWER!!!
d. 2×2 atoms of Al ≠ 1×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
0.2 is the value of coefficient of friction (k)
F=kN
F=horizontal force
n=Normal Force
k=coefficient of friction
k=F/N
k=200/1000
k=0.2
The ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them is known as the friction coefficient. Usually, the Greek letter mu is used to indicate it .N is the normal force, and F is the frictional force, hence F = N/N.
Due to the fact that both F and N are measured in units of force, the coefficient of friction has no dimensions (such as newtons or pounds). The coefficient of friction can have a variety of values for both static and dynamic friction. Static friction occurs when an object encounters friction that resists any applied force, keeping the object at rest until the static frictional force is released. In kinetic friction, the frictional force resists the motion of the object.
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Answer:
When the speed is changing is the correct answer
Explanation:
Because acceleration is rate of change of velocity
PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE
Answer:
a) 250 N/m
b) 22.4 rad/s , 3.6 Hz , 0.28 sec
c) 0.3125 J
Explanation:
a)
F = force applied on the spring = 7.50 N
x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m
k = spring constant of the spring
Since the force applied causes the spring to stretch
F = k x
7.50 = k (0.03)
k = 250 N/m
b)
m = mass of the particle attached to the spring = 0.500 kg
Angular frequency of motion is given as
= 22.4 rad/s
= frequency
Angular frequency is also given as
= 2 π
22.4 = 2 (3.14) f
= 3.6 Hz
= Time period
Time period is given as
= 0.28 sec
c)
A = amplitude of motion = 5 cm = 0.05 m
Total energy of the spring-block system is given as
U = (0.5) k A²
U = (0.5) (250) (0.05)²
U = 0.3125 J
