What is one characteristic of a sample of matter that indicates it is a pure substance? A) It can be separated using by a physic
2 answers:
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Answer:
c. natural force or pull toward the earth
Explanation:
Gravity of the earth is the force of attraction that it naturally possesses to attract any mass.
An apple falls on the earth due to this force of gravity.
The force of gravity near to the surface of the earth is given as:
where:
m= mass of the body
g= acceleration due to gravity
The variation of the gravitational force with height is given as:
where:
where
R = radius of the earth
Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + + t₁ =6 + 4 + 1.6 = 11.6 seconds.