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2 years ago

ILL GIVE YOU BRAINLIEST PLEASE HELP PLEASE

Engineering

1 answer:

Lesechka [4]2 years ago

4 0

Answer:

try to pop it back in good luck im scared for you

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Acquisition of resources from an external source is called?

densk [106]

Answer:

subcontracting

Explanation:

I hope this is right

6 0

2 years ago

Global Courier Services will ship your package based on how much it weighs and how far you are sending the package. Packages abo

denis23 [38]

Answer:

The code will be:

#include <stdio.h>

#include <stdlib.h>

main () {

double weight, shippingCharge, rate, segments;

int distance;

printf("Enter the weight: \n");

scanf("%lf", &weight);

printf("Enter the distance: \n");

scanf("%i", &distance);

if (weight <= 10) {

printf("Rate is $3.00 \n");

rate = 3;

} else {

printf("Rate is $5.00 \n");

rate = 5;

}

if (distance % 500 == 0) {

segments = distance / 500;

} else {

segments = distance / 500 + 1;

}

shippingCharge = rate * segments;

if (distance >1000) {

shippingCharge = shippingCharge + 10;

}

printf("Your shipping charge is $%lf\n", shippingCharge);

system ("pause");

}

8 0

2 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

5 0

2 years ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0

2 years ago

A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i

Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature $T_{1}$ = 300 F = 422.04 K

Duct temperature $T_{2}$ = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant $\sigma$ = 5.67 × $10^{-8}$ $\frac{W}{m^{2} K^{4} }$

Heat transfer per foot length is given by

Q = ∈ $\sigma$ A ( $T_{1}^{4} - T_{2} ^{4}$ ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × $10^{-8}$ × 3.14 × 0.0508 × 1 × ( $422.04^{4} - 294.26^{4}$ )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0

2 years ago