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Kinetic energy is the energy from movement. Hence the fastest movement is at the bottom of the loop - therefore Kinetic energy is highest at the flat bottom of the loop.
Potential energy is the opposite - it occurs at the top of the loop when the car moves the slowest
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>
Answer:
43.868 J
Explanation:
Kinetic energy of a body is the amount of energy possessed by a moving body. The SI unit of kinetic energy is the joule (kg⋅m²⋅s⁻²).
According to classical mechanics, kinetic energy = 1/2 m·v²
Where, m= mass in kg and v= velocity in m/s
Given: m = 19.2 lb and v = 7.10 miles/h
Since, 1 lb= 0.453592 kg
∴ m = 19.2 lb = 19.2 × 0.453592 kg = 8.709 kg
Also, 1 mi = 1609.34 m and 1 h = 3600 sec
∴ v = 7.10 mi/h = 7.10 × 1609.34 m ÷ 3600 sec = 3.174 m/sec
Therefore, <u>kinetic energy of the goose</u> = 1/2 m·v² = 1/2 × (8.709 kg)× (3.174 m/sec)² = 43.868 J
Answer:
<h2>The answer is 4 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
where
a is the acceleration
f is the force
m is the mass
From the question
f = 20 N
m = 5 kg
We have
We have the final answer as
<h3>4 m/s²</h3>
Hope this helps you
