Answer:
dont have the answer but you cute
Explanation:
Answer:
a) 7200 ft/s²
b) 140 ft
c) 3.7 s
Explanation:
(a) Average acceleration is the change in velocity over change in time.
a_avg = Δv / Δt
We need to find what velocity the puck reached after it was hit by the hockey player.
We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:
v² = v₀² + 2a(x − x₀)
(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)
v₀² = 5200 ft²/s²
v₀ = 20√13 ft/s
So the average acceleration impacted to the puck as it is struck is:
a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)
a_avg = 2000√13 ft/s²
a_avg ≈ 7200 ft/s²
(b) The distance the puck travels before stopping is:
v² = v₀² + 2a(x − x₀)
(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)
x = 140 ft
(c) The time the puck takes to travel 10 ft without friction is:
t = (10 ft) / (20√13 ft/s)
t = (√13)/26 s
The time the puck travels over the rough ice is:
v = at + v₀
(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)
t = √13 s
So the total time is:
t = (√13)/26 s + √13 s
t = (27√13)/26 s
t ≈ 3.7 s
The number of waves arriving at the same place in a fixed amount
of time is directly related to the frequency of the waves.
Bond<span> polarity is </span>important<span> because it helps to determine the polarity of </span>molecules<span> and hence their physical properties.</span>
Answer:
(a) 25 m
(b) 75 m
Explanation:
Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.
So, the speed of the jogger,
Let d be the distance covered by him in time, t s.
As distance=(speed) x (time)
So, 
From equation (i)
As the jogger starts from origin, so, the distance, 
, also represents the position of the jogger at the time 
s.
The position-time graph has been shown.
(a) From equation (ii), for t=5.0 s
So, the jogger is at a distance of 25 m from the origin.
(b) Similarly, for t=15.0 s
So, the jogger is at a distance of 75 m from the origin.
