Answer:
28.23 years
Explanation:
I = 1100 A
L = 230 km = 230, 000 m
diameter = 2 cm
radius, r = 1 cm = 0.01 m
Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2
n = 8.5 x 10^28 per cubic metre
Use the relation
I = n e A vd
vd = I / n e A
vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)
vd = 2.58 x 10^-4 m/s
Let time taken is t.
Distance = velocity x time
t = distance / velocity = L / vd
t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second
t = 28.23 years
Answer:

Explanation:
Given that,
The length of a string, l = 0.87 m
Speed of the ball, v = 3.36 m/s
We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

So, the acceleration of the ball is
.
Answer:
His final velocity is 15.8 m/s.
Step-by-step explanation:
Given:
Initial velocity of the driver is,
m/s
Acceleration of the driver is,
m/s²
Time taken to reach final velocity is,
s.
The final velocity is given using the Newton's equations of motion as:
, where,
is the final velocity.
Now, plug in the given values and solve for
.

Therefore, his final velocity is 15.8 m/s.
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