Answer:
a = -7.29 m / s²
Explanation:
For this exercise we must use Newton's second law,
F -W = m a
Force is electrical force
F = k q₁ q₂ / r²
k q₁ q₂ / r² -mg = m a
indicate that the charge of the two spheres is equal
q₁ = q₂ = q
a = (k q² / r² - m g) / m
a = k q² / m r² - g
Let's reduce the magnitudes to the SI system
m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg
q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C
r = 10.0 cm (1m / 100cm) = 0.1000 m
let's calculate
a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8
a = -7.29 m / s²
The negative sign indicates that the direction of this acceleration is downward
The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.
Answer:
D. At x=0, it's acceleration is at a maximum
Explanation:
As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.
Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.
From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.
The same can be said as the box travels backward from point A to -A
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
The speed of the pin after the elastic collision is 9 m/s east.
<h3>
Final speed of the pin</h3>
The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + mu2 = m1v1 + m2v2
where;
- m is the mass of the objects
- u is the initial speed of the objects
- v is the final speed of the objects
4(1.4) + 0.4(0) = 4(0.5) + 0.4v2
5.6 = 2 + 0.4v2
5.6 - 2 = 0.4v2
3.6 = 0.4v2
v2 = 3.6/0.4
v2 = 9 m/s
Thus, The speed of the pin after the elastic collision is 9 m/s east.
Learn more about linear momentum here: brainly.com/question/7538238
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One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A.
This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself.
