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atroni [7]
3 years ago
15

fire work a travels at the speed 300 ft/s firework b travels 240 ft/s. firework b is launched 0.25s before firework a. how many

seconds after firework b launches will both fireworks explode?
Mathematics
1 answer:
storchak [24]3 years ago
7 0

Answer:

Both fireworks  will explode after 1 seconds after firework b launches.

Step-by-step explanation:

Given:

Speed of fire work A= 300 ft/s

Speed of Firework B=240 ft/s

Time before which fire work b is launched =0.25s

To Find:

How many seconds after firework b launches will both fireworks explode=?

Solution:

Let t be the time(seconds) after which both the fireworks explode.

By the time the firework a has been launched, Firework B has been launch 0.25 s, So we can treat them as two separate equation

Firework A= 330(t)

Firework B=240(t)+240(0.25)

Since we need to know the same time after which they explode, we can equate both the equations

330(t) = 240(t)+240(0.25)

300(t)= 240(t)+60

300(t)-240(t)= 60

60(t)=60

t=\frac{60}{60}

t=1

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The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is:
jolli1 [7]

Answer:

Yes, F is a continuous function of r

Step-by-step explanation:

We are given that

When r<R

F(r)=\frac{GMr}{R^3}

r\geq R

F(r)=\frac{GM}{r^2}

Where M=Mass of the earth

R=Radius of earth

G=Gravitational constant

We have to find the function is continuous of r or not.

LHL

\lim_{r\rightarrow R-}\frac{GMr}{R^3}=\frac{GMR}{R^3}=\frac{GM}{R^2}

RHL

\lim_{r\rightarrow R+}\frac{GM}{R^2}=\frac{GM}{R^2}

F(R)=\frac{GM}{R^2}

When a function is continuous at x=a

Then, LHL=RHL=f(a)

RHL==LHL=F(R)

Hence, the function is continuous of r.

8 0
3 years ago
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Answer:

Step-by-step explanation:

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8 0
3 years ago
Write the computer program for;urgent​
AlladinOne [14]

ANSWER:

#include <stdio.h>

#define PAYRATE 10 //basic pay rate per hour

#define OVERTIME 15 //in excess of 40 hours a week

int NetPay(int hours);

int main()

{

int userWeeklyHours;

printf("Please enter your total weekly working hours: \n");

scanf("%d", &userWeeklyHours);// get the user weekly hours

NetPay(userWeeklyHours);

return 0;

}

int NetPay(int hours)// implementing the function to calculate total pay, total taxes, and net pay

{

int firstRate, secondRate, restOfRate, secondAmount, rest, payAfterTax, payedBeforeTax, overHours;

if (hours > 40)

{

overHours = hours - 40;

payedBeforeTax = (40 * PAYRATE) + (overHours * OVERTIME);

}

else

payedBeforeTax = hours * PAYRATE;//defining the user first paycheck before taxes

if (payedBeforeTax <= 300)//paying only first rate case

{

firstRate = payedBeforeTax*0.15;

payAfterTax =payedBeforeTax - firstRate;

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate, payAfterTax);

}

else if (payedBeforeTax > 300 && payedBeforeTax <= 450)//paying first and second rate

{

secondAmount = payedBeforeTax - 300;

firstRate = (payedBeforeTax - secondAmount) * 0.15;

secondRate = secondAmount * 0.20;

payAfterTax = payedBeforeTax - (firstRate + secondRate);

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate + secondRate, payAfterTax);

}

else if (payedBeforeTax > 450)// paying all rates

{

rest = payedBeforeTax - 450;

secondAmount = (payedBeforeTax - 300) - rest;

firstRate = (payedBeforeTax - (rest + secondAmount)) * 0.15;

secondRate = secondAmount * 0.20;

restOfRate = rest * 0.25;

payAfterTax = payedBeforeTax - (firstRate + secondRate + restOfRate);

printf("your total gross is %d, your taxes to pay are %d, your net pay is %d", payedBeforeTax, firstRate + secondRate + restOfRate, payAfterTax);

}

return payAfterTax;

}

8 0
3 years ago
Need Help ASAP !!!!!!
lara31 [8.8K]

if you in you computer you can use math/way

3 0
2 years ago
Read 2 more answers
Don't know the claim or how to explain
Marrrta [24]
Is this calculus or something?
7 0
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