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aleksley [76]
3 years ago
15

Convert: 410 km = ________ mi

Mathematics
2 answers:
vampirchik [111]3 years ago
7 0
<span>410 km = 254.762 mi

HOPE THIS HELPS! ^_^</span>
Ilya [14]3 years ago
5 0
254.76218881717  ther is ur answer
You might be interested in
A body of constant mass m is projected vertically upward with an initial velocity v0 in a medium offering a resistance k|v|, whe
bixtya [17]

Answer:

tm = tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] }

Xm = Xₐ = (v₀m)/k - ({m²g}/k²) ㏑(1+{kv₀/mg})

Step-by-step explanation:

Note, I substituted maximum time tm = tₐ and maximum height Xm = Xₐ

We will use linear ordinary differential equation (ODE) to solve this question.

Remember that Force F = ma in 2nd Newton law, where m is mass and a is acceleration

Acceleration a is also the rate of change in velocity per time. i.e a=dv/dt

Therefore F = m(dv/dt) = m (v₂-v₁)/t

There are two forces involved in this situation which are F₁ and F₂, where F₁ is the gravitational force and F₂ is the air resistance force.

Then, F = F₁ + F₂ = m (v₂-v₁)/t

F₁ + F₂ = -mg-kv = m (v₂-v₁)/t

Divide through by m to get

-g-(kv/m) = (v₂-v₁)/t

Let (v₂-v₁)/t be v¹

Therefore, -g-(kv/m) = v¹

-g = v¹ + (k/m)v --------------------------------------------------(i)

Equation (i) is a inhomogenous linear ordinary differential equation (ODE)

Therefore let A(t) = k/m and B(t) = -g --------------------------------(ia)

b = ∫Adt

Since A = k/m, then

b = ∫(k/m)dt

The integral will give us b = kt/m------------------------------------(ii)

The integrating factor will be eᵇ = e ⁽<em>k/m</em>⁾

The general solution of velocity at any given time is

v(t) = e⁻⁽b⁾ [ c + ∫Beᵇdt ] --------------------------------------(iiI)

substitute the values of b, eᵇ, and B into equation (iii)

v(t) = e⁻⁽kt/m⁾ [ c + ∫₋g e⁽kt/m⁾dt ]

Integrating and cancelling the bracket, we get

v(t) = ce⁻⁽kt/m⁾ + (e⁻⁽kt/m⁾ ∫₋g e⁽kt/m⁾dt ])

v(t) = ce⁻⁽kt/m⁾ - e⁻⁽kt/m⁾ ∫g e⁽kt/m⁾dt ]

v(t) = ce⁻⁽kt/m⁾ -mg/k -------------------------------------------------------(iv)

Note that at initial velocity v₀, time t is 0, therefore v₀ = v(t)

v₀ = V(t) = V(0)

substitute t = 0 in equation (iv)

v₀ = ce⁻⁽k0/m⁾ -mg/k

v₀ = c(1) -mg/k = c - mg/k

Therefore c = v₀ + mg/k  ------------------------------------------------(v)

Substitute equation (v) into (iv)

v(t) = [v₀ + mg/k] e⁻⁽kt/m⁾ - mg/k ----------------------------------------(vi)

Now at maximum height Xₐ, the time will be tₐ

Now change V(t) as V(tₐ) and equate it to 0 to get the maximum time tₐ.

v(t) = v(tₐ) = [v₀ + mg/k] e⁻⁽ktₐ/m⁾ - mg/k = 0

to find tₐ from the equation,

[v₀ + mg/k] e⁻⁽ktₐ/m⁾ = mg/k

e⁻⁽ktₐ/m⁾ = {mg/k] / [v₀ + mg/k]

-ktₐ/m = ㏑{ [mg/k] / [v₀ + mg/k] }

-ktₐ = m ㏑{ [mg/k] / [v₀ + mg/k] }

tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] }

Therefore tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] } ----------------------------------(A)

we can also write equ (A) as tₐ = m/k ㏑{ [mg/k] [v₀ + mg/k] } due to the negative sign coming together with the In sign.

Now to find the maximum height Xₐ, the equation must be written in terms of v and x.

This means dv/dt = v(dv/dx) ---------------------------------------(vii)

Remember equation (i) above  -g = v¹ + (k/m)v

Given that dv/dt = v¹

and -g-(kv/m) = v¹

Therefore subt v¹ into equ (vii) above to get

-g-(kv/m) = v(dv/dx)

Divide through by v to get

[-g-(kv/m)] / v = dv / dx -----------------------------------------------(viii)

Expand the LEFT hand size more to get

[-g-(kv/m)] / v = - (k/m) / [1 - { mg/k) / (mg/k + v) } ] ---------------------(ix)

Now substitute equ (ix) in equ (viii)

- (k/m) / [1 - { mg/k) / (mg/k + v) } ] = dv / dx

Cross-multify the equation to get

- (k/m) dx = [1 - { mg/k) / (mg/k + v) } ] dv --------------------------------(x)

Remember that at maximum height, t = 0, then x = 0

t = tₐ and X = Xₐ

Then integrate the left and right side of equation (x) from v₀ to 0 and 0 to Xₐ respectively to get:

-v₀ + (mg/k) ㏑v₀ = - {k/m} Xₐ

Divide through by - {k/m} to get

Xₐ = -v₀ + (mg/k) ㏑v₀ / (- {k/m})

Xₐ = {m/k}v₀ - {m²g}/k² ㏑(1+{kv₀/mg})

Therefore Xₐ = (v₀m)/k - ({m²g}/k²) ㏑(1+{kv₀/mg}) ---------------------------(B)

3 0
3 years ago
What is the range of the relation?<br> (-3,-2,0,2)<br> (-3,3)<br> (-4,-2,1,2)<br> (-4,-3,-2,1,0,2)
9966 [12]

Answer:

jn8huuojgui

Step-by-s

3 0
2 years ago
Write an iequality for -5(×+7)&lt;-10 representing the solution for x​
vovikov84 [41]

Answer:

X<-5

Step-by-step explanation:

1) open the brackets

2) take -35 to the other side with -10

where it turns to be -5X<-10+35

3) work out to find -5X<25

4) divide both sides by -5

to find X<-5

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%3Cbr%3E%0Ay%3D%20x%5E%7B2%7D%20-9" id="TexFormula1" title="&amp;#10;y= x^{2} -9" alt="&amp;#1
svp [43]
Simple...

y=x^{2} -9

y= x^{2} -9

It's quite easy to see x^{2} = x^{2}

Thus, any value of x makes the equation true.

Thus, your answer.
7 0
3 years ago
Use one of the triangles to approximate PQ in the triangle below.
s2008m [1.1K]

Answer:

The answer is B 5.4 units

Step-by-step explanation:

4 0
2 years ago
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