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HACTEHA [7]
1 year ago
11

A boat sails 285 miles south and then 132 miles west. What is the magnitude of the boats resultant vector?

Mathematics
1 answer:
Delicious77 [7]1 year ago
8 0

The magnitude of the boats resultant vector is 314.1 mi

<h3>What is a vector?</h3>

A vector is a physical quantity that has both magnitude and direction.

<h3>What is a resultant vector?</h3>

A resultant vector is the sum of two or more vectors.

<h3>How to find the boats resultant vector?</h3>

Since the boat sails 285 miles south and then 132 miles west, we have that its first direction vector is r = (285 mi)j. Also, its direction vector west is r' = -(132 mi)i

So, the resultant vector R = r + r'

=  (285 mi)j + (132 mi)i

=  (132 mi)i + (285 mi)j

So, the magnitude of the resultant vector is R = √(r² + r'²)

So, substituting thevalues of the variables into the equation, we have

R = √(r² + r'²)

R = √((285 mi)² + (132 mi)²)

R = √(81225 mi² + 17424 mi²)

R = √(98649 mi²)

R = 314.08 mi

R ≅ 314.1 mi

So, the resultant vector is 314.1 mi

Learn more about magnitude of resultant vector here:

brainly.com/question/28047791

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masya89 [10]

Answer:

Question not complete,

So i will analyse the possible problem

Step-by-step explanation:

A tank contain 2200L

Volume V = 2200L

Solution of 0.06kg/L of sugar

Rate of entry i.e input

dL/dt=5L/min

Let y(t) be the amount of sugar in tank at any time.

But at the beginning there was no sugar in the tank

i.e, y(0)=0, this will be out initial value problem,

The rate of amount of sugar at anytime t is

dy/dt=input amount of sugar - output amount of sugar.

Now,

Then rate of input is

5L/min × 0.06kg/L

Then, input rate= 0.3kg/min

Output rate is

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then, output rate = y(t)/440 kg/min

Now then,

dy/dt=input rate -output rate

dy/dt=0.3-y/440

Cross multiply through by 400

400dy/dt=120-y

Using variable separation

400/(120-y) dy = dt

∫400/(120-y) dy = ∫dt

-400In(120-y)=t +C

In(120-y)=-t/400+C/400

C/400 is another constant, let say B

In(120-y)=-t/400+B

Take exponential of both side

120-y(t)=exp(-t/400+B)

120-y(t)=exp(-t/400)exp(B)

exp(B) is a constant let say C

-y(t)=Cexp(-t/400)-120

y(t)=120-Cexp(t/400)

Now, the initial condition

a. At the start the mass of sugar in the water is 0 because it is just pure water at start.

Therefore y(0)=0,

b. Applying this to y(t)

y(t)=120-Cexp(-t/400)

y=0, t=0

0=120-Cexp(0)

0=120-C

C=120

Therefore,

y(t)=120 - 120exp(-t/400)

Let know the mass rate as t tends to infinity

At infinity

exp(-∞)=1/exp(∞)=1/∞=0

Then,

The exponential aspect tend to 0

Then, y(t)=120 as t tend to ∞

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