Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.
For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.
If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.
Good morning, Jesusperez7!
Semi-conductors are materials that have insulator and conductor properties.
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Answer:
t = 1.75
t = 0.04
Explanation:
a)
For part 1 we want to use a kenamatic equation with constant acceleration:
X = 1/2*a*t^2
isolate time
t = sqrt(2X / a)
Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2
t = sqrt(2*15m / 9.8m/s^2)
t = 1.75 s
b)
The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:
v = d / t
isolate time
t = d / v
plug in known variables
t = 15m / 340m/s
t = 0.04 s
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
To calculate the change in kinetic energy, you must know the force as a function of position. The work done by the force causes the kinetic energy change
Explanation:
The work-energy theorem states that the change in kinetic enegy of an object is equal to the work done on the object:
where the work done is the integral of the force over the position of the object:
As we see from the formula, the magnitude of the force F(x) can be dependent from the position of the object, therefore in order to solve correctly the integral and find the work done on the object, it is required to know the behaviour of the force as a function of the position, x.
