A force is a push or pull acting upon an object as a result of its interaction with another object. There are a variety of types of forces. a variety of force types were placed into two broad category headings on the basis of whether the force resulted from the contact or non-contact of the two interacting objects.
Contact Forces
Action-at-a-Distance Forces
Frictional Force
Gravitational Force
Tensional Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force
These are types of individual forces
Applied Force
Gravitational Force
Normal Force
Frictional Force
Air Resistance Force
Tensional Force
Spring Force
Answer:
What is a Pure Substance?
Explanation:
It is something which cannot be divided into parts by physical means, as it's all made up of the same thing. Pure substances are either elements or compounds. Elements can NOT be separated into other types of matter (physically or chemically).
Answer:
increase
Explanation:
think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing
Answer:
m = 1 kg
Explanation:
Given that,
The force constant of the spring, k = 39.5 N/m
The frequency of oscillation, f = 1 Hz
The frequency of oscillation is given by the formula as formula as follows :
So, the mass that is attached to the spring is 1 kg.
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
