Ask question

Ask question

All categories

2 years ago

3

Write three statements to print the first three elements of array runTimes. Follow each statement with a newline. Ex: If runTime

s[5] = {800, 775, 790, 805, 808}, print: 800 775 790

1 answer:

Dmitry_Shevchenko [17]2 years ago

6 0

Answer:

runTimes = [800, 775, 790, 805, 808]

x = runTimes[0:3]

print(x)

Explanation:

The code is written in python . The question asked us to print the first elements of an array or list. The first three statement of a list can be printed by using the index of the list.

runTimes = [800, 775, 790, 805, 808]

This is the list or array containing elements .The variable runTimes is used to store the array or list.

x = runTimes[0:3]

The variable x is used to store the code, runTimes[0:3] . The code means the first three elements of the array or list. From index 0 to 3 .

print(x)

The print function is now used to display the first three elements of the array.

You might be interested in

To water his lawn, a homeowner uses two hoses. One connects to the faucet, the other to the end of the first hose to make the ho

Shtirlitz [24]

Answer: to be exact you need 28mm of tubing for that

Explanation:

When the election

8 0

2 years ago

The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc

Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑ $f_{y}$ = 0, ∑M = 0

If ∑ $M_{A}$ = 0

⇒ $F_{BC}$ ×0.9 - P × 0.6 = 0

⇒ $F_{BC}$ ×3 - P × 2 = 0

⇒ $F_{BC}$ = $\frac{2P}{3}$

If ∑ $M_{B}$ = 0

⇒ $F_{AD}$ ×0.9 = P × 0.3

⇒ $F_{AD}$ ×3 = P

⇒ $F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒ $\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0

2 years ago

Q9. A cylindrical specimen of a metal alloy 54.8 mm long and 10.8 mm in diameter is stressed in tension. A true stress of 365 MP

wolverine [178]

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

<u>σ = 391.2 MPa</u>

7 0

2 years ago

A cable consists of a steel wire 2.5mm in diameter, surrounded by six bronze wires of the same size. If the allowable stress in

DENIUS [597]

Answer:

Explanation:

given data:

steel diamter = 25 mm

allowable stress in bronze = 60 MPa

length of wire = 30 m

Ebr=80GPa

Est = 200GPA

AREA of wire $= \frac{\pi}{4} 2.5^2 = 4.906 mm2$

total load = 6*Pbr + Pst .............1

where

Pbr is load due to bronze

Psr is load due to steel

elongationis given as

$\frac{Pbr*l}{Ebr*A} =\frac{Pst*l}{Est*A}$

$Pbr = \frac{ Ebr}{Est} Pst$

$= \frac{ 80}{200}Pst$

force on bronze wire

$\sigma br = \frac{Pbr}{A}$

$60 = \frac{Pbr}{4.90625}$

Pbr = 294.375 N

from eqt 2

$294.375 = \frac{80}{200} Pst$

Pst = 735.93 N

SAFE LOAD from eq 1

P = 6 *294.375 + 735.937 = 2502.18 N

EXtension under load

$\delta t = \frac{Pbr L}{AEbr} = \frac{294.37*30*10^3}{4.90625*80*10^3}$

$\delta t = 22.5 mm$

6 0

2 years ago

A golfer and her caddy see lightning nearby. the golfer is about to take his shot with a metal club, while her caddy is holding

Sveta_85 [38]

Answer:

The golfer is at greater risk.

Explanation:

The golfer is holding a metal club. Metal is a good conductor for electricity (lightning), meaning electrons can pass through easily. Her caddy is at lesser risk because she is holding a plastic handled umbrella. Plastic is an insulator, which does not easily allow the movement of electrons to pass.

7 0

2 years ago