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Elena L [17]
3 years ago
7

Basic C++ For Loop I'm trying to learn. Replit tells me that the for in the forloop is an error, but I don't know what's wrong.

Can anyone help?

Engineering
1 answer:
STatiana [176]3 years ago
8 0

Answer:

  curly brackets are missing

Explanation:

The body of the main() function need to be enclosed in curly brackets. Try this:

int P = 3000;

int main( ) {

  for (int t = 0; t < 10; t++) {

     cout << P;

  }

}

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A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
UkoKoshka [18]
Maybe it willl be I don’t know
7 0
3 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density
pav-90 [236]
Here is the flow sheet. Hope this helps have a great day!!

3 0
3 years ago
The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find
PSYCHO15rus [73]

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 %  binder content 2.495

Therefore

95 % mix + 5 % binder  gives S.G. = 2.495

Where the  binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

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Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

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Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

3 0
3 years ago
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