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2 years ago

You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?

Engineering

1 answer:

nalin [4]2 years ago

8 0

Answer:

please give brainliest my brother just got the corona virus

Explanation:

this is my brothers account he wants to get 5 brainliest

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CHALLENGE ACTIVITY 2.8.1: Using constants in expressions. The cost to ship a package is a flat fee of 75 cents plus 25 cents per

mihalych1998 [28]

Answer:

Weight(lb): 10

Flat fee(cents): 75

Cents per pound: 25

Shipping cost(cents): 325

Explanation:

we run this as a jave programming language

import java.util.Scanner;

public class Shipping Calculator {

public static void main (String [] args) {

int shipWeightPounds = 10;

int shipCostCents = 0;

final int FLAT_FEE_CENTS = 75;

final int CENTS_PER_POUND = 25;

shipCostCents = FLAT_FEE_CENTS + CENTS_PER_POUND * shipWeightPound

/* look up the solutioin above */

System.out.println("Weight(lb): " + shipWeightPounds);

System.out.println("Flat fee(cents): " + FLAT_FEE_CENTS);

System.out.println("Cents per pound: " + CENTS_PER_POUND);

System.out.println("Shipping cost(cents): " + shipCostCents);

}

7 0

2 years ago

Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA

weqwewe [10]

OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.

It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.

5 0

2 years ago

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

2 years ago

The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13

prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0

2 years ago

An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

$F = LwY_{avg}$

where;

w is the width of the annealed copper

$Y_{avg}$ is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

$L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm$

Now, determine the average true stress, $Y_{avg}$ , for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

$F = LwY_{avg}$

$F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN$

The power required in this operation is given by;

$P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW$

5 0

2 years ago