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babymother [125]

3 years ago

11

If you have 300 skittles in a bag and you need to have 28 percent yellow. How many yellow skittles would you have to make a mini

mum of 28 percent in the bag?

1 answer:

masha68 [24]3 years ago

6 0

Answer:

im trying to do a challenge because this kid deleted all my answers

Explanation:

You might be interested in

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

If my friend have the corona what do I do

nordsb [41]

(E. Call the hospital to take them away

5 0

3 years ago

Read 2 more answers

Open the"stateData3.c" program and try to understand how the tokenization works. If you open the input file "stateData.txt", you

babymother [125]

Answer:

Kindly see explaination

Explanation:

Code

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#define size 200

int main(void)

{

int const numStates = 50;

char tempBuffer[size];

char tmp[size];

char fileName[] = "stateData.txt"; // Name of the text file (input file) which contains states and its populations

char outFile[] = "stateDataOutput1.txt"; // Output file name

// Open the input file, quit if it fails...

FILE *instream = fopen(fileName, "r");

/* Output File variable */

FILE *opstream;

if(instream == NULL) {

fprintf(stderr, "Unable to open file: %s\n", fileName);

exit(1);

}

//TODO: Open the output file in write ("w") mode

/* Opening output file in write mode */

opstream = fopen(outFile, "w");

//TODO: Read the file, line by line and write each line into the output file

//Reading data from file

while(fgets(tmp, size, instream) != NULL)

{

//Writing data to file

fputs(tmp, opstream);

}

// Close the input file

fclose(instream);

//TODO: Close the output file

/* Closing output file */

fclose(opstream);

return 0;

}

5 0

3 years ago

4.68 Steam enters a turbine in a vapor power plant operating at steady state at 560°C, 80 bar, and exits as a saturated vapor at

garik1379 [7]

Answer:

please mark me as a brainleast

Explanation:

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3 0

2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago