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Galina-37 [17]
3 years ago
8

If you need to write a function that will compute the cost of some candy, where each piece costs 25 cents, which would be an app

ropriate function declaration?
(A) int calculateCost(char name);
(B) char calculateCost(int count);
(C) int calculateCost int count;
(D) int calculateCost(int count);
Engineering
1 answer:
masya89 [10]3 years ago
6 0
The best answer would be

D. Int calculateCost(int count);
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IF A CAR AHEAD OF YOU HAS STOPPED AT A CROSSWALK, YOU SHOULD:
Vsevolod [243]

Answer:

well, depend if theres a stop sign or a light signal..you can always just honk at him/her so they proceed

Explanation:

but otherwise, the answer would be

A: stop and proceed when safe

<em><u></u></em>

<em><u>Hope this helped! Have a nice day, and feel free to follow me on insta (leo_g_rios)</u></em>

<em><u>-XxDeathshotxX</u></em>

3 0
3 years ago
Lately, you have noticed some repetitive stress in your wrist. Which sign is most likely the cause of that stress and pain?
Cerrena [4.2K]
1, you might have been carrying things that are way too heavy for you.
2, you might have weak tendons.
3 0
3 years ago
Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c
Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress  acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

  = 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

    = (1.8225*10⁻⁷/1.14) * (35000)

    = 1.60*10⁻⁷ * 35000

   = 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ  = h*(Δp/L)

    = 0.0045* 35000

    = 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

   = (0.0045²/2*0.38) * 35000

   =2.664*10⁻⁵ *35000

   = 0.93 m/s

7 0
3 years ago
: A drive system using the electric motor is under load as 75Nm with an angular velocity of 100rad/s, then the electric motor is
ahrayia [7]

Answer:unsure about how long.

Explanation:

3 0
3 years ago
A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the
tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress  = 50 MPa

amplitude stress  = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first  maximum stress  and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

\sigma _{m} =  \sigma _{maximum} + \sigma _{minimum}  / 2

50 =  \sigma _{maximum} + \sigma _{minimum}  / 2    ...........1

and

225 =  \sigma _{maximum} + \sigma _{minimum}  / 2      ...........2

from eqution 1 and 2 we get maximum and minimum stress

\sigma _{maximum} = 275 MPa        ............3

and \sigma _{minimum} = -175 MPa     ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio =  \sigma _{minimum} / \sigma _{maximum}

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = \sigma _{maximum} - \sigma _{minimum}  

magnitude = 275 - (-175) = 450 MPa

3 0
3 years ago
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