Answer:
Explanation:
Read forces directly from the graph.
Read displacements directly from the graph.
Use the area under the graph to find the work done by the force. This is equal to the kinetic or potential energy the object gains due to the application of the force.
The equilibrium condition allows finding the result for the force that the chair exerts on the student is:
- The reaction force that the chair exerts on the student's support is equal to the student's weight.
Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.
∑ F = 0
Where F is the external force.
The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.
Let's analyze the force on the chair.
Let's analyze the forces on the student.
In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:
- The reaction force that the chair exerts on the student's support is equal to the student's weight.
Learn more here: brainly.com/question/18117041
When two atoms of the same nonmetal react,they form what we know today as a diatomic molecule.
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Answer:
Vf = 4.77 m/s
Explanation:
During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height = 1.16 m
Vf = Final Velocity of Ball = ?
Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)
Therefore, using these values in the equation, we get:
(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²
Vf = √(22.736 m²/s²)
<u>Vf = 4.77 m/s</u>
Answer:
x(t) = ⅟₁₀₈t⁴ + 10t + 24
v(t) = ⅟₂₇t³ + 10
Explanation:
a(t) = C₁t²
velocity is the integral of acceleration
v(t) = ⅓C₁t³ + C₂
position is the integral of velocity
x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃
x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃
C₃ = 24
x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24
72 = 108C₁ + 6C₂
C₂ = 12 - 18C₁
v(6) = 18 = ⅓C₁6³ + C₂
18 = 72C₁ + C₂
18 = 72C₁ + (12 - 18C₁)
6 = 54C₁
C₁ = 1/9
C₂ = 12 - 18(1/9)
C₂ = 10
