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Oduvanchick [21]
3 years ago
9

Consider the following list of numbers. 129, 685, 125, 511, 601, 52, 46 The height of a binary search tree is the maximum number

of edges you have to go through to reach the bottom of the tree, starting at the root. What is the height of the tree for the numbers above, in the order given?
Physics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

The height of the tree is three (3) deep

Explanation:

It's 3 deep

Under 129, comes 125 and 685;

Under 125, comes 52 : Under 685, comes 511;

Under 52, comes 46 : Under 511, is 601.

This is illustrated below.

129

∧

125,685

|,|

52,511

|,|

46,601

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The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing
Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0
3 years ago
If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o
larisa86 [58]

Answer:

f_k = 310N

the answer is A.

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.

Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:

∑F = m(0)

∑F = 0

It means that:

F -f_k = 0

where F is the force applied and f_k is the friction force. Replacing the value of F, we get:

310N -f_k = 0

Finally, solving for f_k:

f_k = 310N

8 0
3 years ago
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
2 years ago
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seraphim [82]

Try this option, the answers are marked with colour.

8 0
3 years ago
A student wants to design an experiment to study the transformation of mechanical energy. Which object can be used to investigat
Vsevolod [243]
I guess it’s the 3rd one
7 0
2 years ago
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