A 0.25-kilogram ball is observed to accelerate at 4,000 m/sec2 as it is hit with a bat.
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<h2>Hello!</h2>
The answer is: 201 J of kinetic energy.
<h2>Why?</h2>The motion of a pendulum (with no friction considered) is a continuous exchange between potential energy and kinetic energy.
So, at the highest point of its swing, the potential energy will be the maximum potential energy that the pendulum can have and the kinetic energy will be 0 since at max height the speed tends to 0.
On the opposite side, when the pendulum is at the bottom (the lowest point of its swing) the potential energy will be the minimum (tends to 0) but the kinetic energy will be the maximum.
Also, in the pendulum motion, the total energy is conserved, meaning that:
Where,
PE(h),is the potential energy at the highest point.
KE(h), is the kinetic energy at the highest point.
PE(l), is the potential energy at the lowest point (bottom of pendulum swing).
KE(l), is the kinetic energy at the lowest point (bottom of pendulum swing).
m, is the mass of the object.
g, is the acceleration of gravity.
v, is the speed of the object.
So, what is the energy at the bottom of its swing?
So, the pendulum has 201 of kinetic energy at the bottom of its swing.
Meaning that the energy is conserved.
Have a nice day!
Answer:
Fn₃= -28.3*10⁻⁶N (in direction -x) :net force exerted by two charges q₁,q₂ on a third charge q₃.
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces .
Equivalences
1nC= 10⁻⁹ C
Known data
k= 9 *10⁹ N*m² /C²
q₁=-15.0 nC=-15*10⁻⁹ C
q₂=+38 nC =+38*10⁻⁹ C
q₃= +46 nC= =+46*10⁻⁹ C
d₁₃= 0.65 m
d₂₃= 1.075 m
d₁₃: distance from q₁ to q3
d₂₃: distance from q₂ to q3
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force F₂₃ of q₂ on q₃ is repulsive because the charges have equal signs and the forces : Force F₂₃ is directed to the left (-x).
The force F₁₃ of q₁ on q₃ is attractive because the charges have opposite signs. Force F₁₃ is directed to the left (-x)
Calculation of the net force exerted for q₁ and q₂ on the charge q₃
The net force exerted for q₁ and q₂ on the charge q₃ is the algebraic sum of the forces F₁₁₃ and F₂₃ because both acts on the x-axis.
Fn₃= F₁₃+F₂₃
To calculate the magnitudes of the forces exerted by the charges q₁, and q₂ on q₃ we apply Coulomb's law:
F₁₃=(k*q₁*q₃)/d₁₃²
F₁₃=(9*10⁹*15*10⁻⁹*46*10⁻⁹)/(0.65)² = 14698.2*10⁻⁹ N = 14.7*10⁻⁶N
F₁₃= 14.7*10⁻⁶N, in the direction of the x-axis negative (-x)
F₃=(k*q₂*q₃)/d₂₃²
F₂₃=(9*10⁹*38*10⁻⁹*46*10⁻⁹)/(1.075)² =13623.4*10⁻⁹ N= 13.6*10⁻⁶N
F₂₃=13.6*10⁻⁶N (in direction -x)
Net force on q₃
Fn₃= -14.7*10⁻⁶N - 13.6*10⁻⁶N = -28.3*10⁻⁶N
Fn₃= -28.3*10⁻⁶N (in direction -x)
