Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, 
rev/sec
rad/sec
= 5.89 rad/sec
Therefore, force required,
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
Explanation:
Given Data
Total mass=93.5 kg
Rock mass=0.310 kg
Initially wagon speed=0.540 m/s
rock speed=16.5 m/s
To Find
The speed of the wagon
Solution
As the wagon rolls, momentum is given as
P=mv
where
m is mass
v is speed
put the values
P=93.5kg × 0.540 m/s
P =50.49 kg×m/s
Now we have to find the momentum of rock
momentum of rock = mv
momentum of rock = (0.310kg)×(16.5 m/s)
momentum of rock =5.115 kg×m/s
From the conservation of momentum we can find the wagons momentum So
wagon momentum=50.49 -5.115 = 45.375 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
Speed of wagon=45.375/(93.5-0.310)
Speed of wagon= 0.487 m/s
Throwing rock backward,
momentum of wagon = 50.49+5.115 = 55.605 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)
speed of wagon= 0.5967 m/s
Answer:
dart
Explanation:
dart and sun and water so that the plant be okay
Answer:
λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
= - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
E = h f
The speed of light is
c = λ f
E = h c /λ
For a transition between two states we have
- 
= - k²e² / 2m (1 / 
² -1 / 
²)
h c / λ = -k² e² / 2m (1 / ² - 1/ ²)
1 / λ = (- k² e² / 2m h c) (1 / ² - 1/²)
The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
1 /λ = 1.097 10⁷ (1/10² - 1/8²)
1 / λ = 1.097 10⁷ (0.01 - 0.015625)
1 /λ = 0.006170625 10⁷
λ = 162 10⁻⁷ m
