How does the vertical acceleration at point a compare to the vertical acceleration at point c?
2 answers:
Both accelerations are equal to free-fall acceleration.
<h3>Further explanation</h3>- This is a matter of projectile motion, i.e., an object moving in two dimensions. The shaped is called a parabola.
- The only forces acting during its flight are gravity and friction. In many situations, air resistance can be ignored.
- It is moving horizontally and vertically at the same time but recalls this, the horizontal dan vertical components of the motion are independent of one another. Assuming the gravitational force is constant.
<u>Horizontal component</u>
There are no forces in the horizontal direction, so there is no horizontal acceleration, i.e., a = 0 m/s². This means the horizontal velocity must be constant along the track.
<u>Vertical component</u>
There is naturally a constant vertical force acting down, so there is a constant vertical acceleration. The value of vertical acceleration is similar to gravity (g).
See the attachment for the motion diagram.
Therefore the vertical acceleration at point A and point B is the same as free-fall acceleration.
<h3>Learn more</h3>- Determine the acceleration of the stuffed bear brainly.com/question/6268248
- Particle's speed and direction of motion brainly.com/question/2814900
- The energy density of the stored energy brainly.com/question/9617400
Keywords: the motion diagram, showing the trajectory, projectile motion, parabola, horizontal, vertical component, gravity, how does the vertical acceleration at point A, compare, point C, equal to free-fall acceleration
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Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.