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 first thing you should know in this case is the following definition:
PV = nRT
Then, as the temperature is constant, then:
PV = k
Then, we have two states:
P1V1 = k
P2V2 = k
We can then equalize both equations:
P1V1 = P2V2
Substituting the values:
(1.25) * (101) = (2.25) * (P2)
Clearing P2:
P2 = ((1.25) * (101)) /(2.25)=56.11Kpa
answer:
the new pressure inside the jar is 56.11Kpa
Answer:
g = 0.85 m
Explanation:
g = 
were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 
N
), M is the mass of the earth (5.972 x 
kg), and h is the distance of meteoroid to the earth.
h = 3.40 x R
= 3.40 x 6371 km
h = 21661.4 km
= 21661400 m
Thus,
g = 
= 
= 0.84944
g = 0.85 m
The acceleration due to the Earth's gravitation is 0.85 m.
Answer:
42244138.951 m
Explanation:
G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²
r = Radius of orbit from center of earth
M = Mass of Earth = 5.98 × 10²⁴ kg
m = Mass of Satellite
The satellite revolves around the Earth at a constant speed
Speed = Distance / Time
The distance is the perimeter of the orbit
The Centripetal force of the satellite is balanced by the universal gravitational force
The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m
