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
Answer:

Explanation:
Here we know that for the given system of charge we have no loss of energy as there is no friction force on it
So we will have


now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.
So we have



so distance moved by the particle is given as



Heat used by electric heater :
Q = m • c • ∆T
Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)
Q = 8.82 × 10⁶ J
Cost of electrical energy :
Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)
Cost = $ 0.3675
We know that
Distance = speed x time
Let w be the time Brad spent walking. The time spent jogging will be 1 - w
6 = 5w + 9(1 - w)
w = 0.75 hours
Distance walked = 0.75 x 5
= 3.75 km
To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as

The given data are given with respect to known constants, for example the mass of the sun is

The radius between the earth and the sun is given by

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:


Substituting in Kepler's third law:






Therefore the period of this star is 3.8years