Explanation:
For projectile motion, use constant acceleration equation:
Δx = v₀ t + ½ at²
where Δx is the displacement,
v₀ is the initial velocity,
a is the acceleration,
and t is time.
Both objects are projected upward with velocity u. The second object is thrown after a time t₀.
For the first object:
Δx = u t + ½ (-g) t²
Δx = u t − ½g t²
For the second object:
Δx = u (t−t₀) + ½ (-g) (t−t₀)²
Δx = u (t−t₀) − ½g (t−t₀)²
Assuming the objects meet, the displacements will be equal:
u t − ½g t² = u (t−t₀) − ½g (t−t₀)²
u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)
u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²
0 = -u t₀ + g tt₀ − ½g t₀²
0 = -u + g t − ½g t₀
g t = u + ½g t₀
t = u/g + t₀/2
I think it’s A that’s the right answer
we can reduce the friction between its moving parts.
Answer:
1.58 W
Explanation:
Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is
From the intensity of the sound we can calculate the power at 12 m
