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galben [10]
3 years ago
7

Please heeeeeeeeeeeeeeeeeeeeeeeeeelp

Mathematics
1 answer:
hram777 [196]3 years ago
5 0
B) x=2 and y=14 i’m pretty sure lol
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
A worm takes 19 seconds to travel 25 cm , How many groups of 25 cm are in 1 meter ? u guys don't need to help me but if u would
Sedbober [7]

Answer:

Step-by-step explanation:

we know that 100 cm are equal to 1 meter

there are 4 "groups" of 25 cm in one meter

the worm needs 76 seconds to travel a meter, in case you need to know that

6 0
3 years ago
The relationship between the number of hours a plumber works and the total work fee she charges is proportional. Her fee for 5 h
dolphi86 [110]

Answer: As given, relationship between the number of hours a plumber works and the total work fee she charges is proportional.

The meaning of proportional is Ratio of number of hours a plumber works and the total work fee she charges is constant.

Using Unitary method

5 hours of work = $350

1 hours of work = $70

Let number of hours worked = x hours

Amount earned for x hours of work is represented by = $ y

For , x hours of work amount earned = 70 x (x=1,2,3...)

→y= 70 x

→→

Step-by-step explanation:

4 0
3 years ago
12-2 one step equations
Grace [21]

Answer:

N + 3 = -5

n= -8

N-4=9

n=13

4+N=10

n=6

-10+N=3

n=13

N-10=1

n=11

-11+n=-12

n=-1

N+19=20

n=1

15+N=-10

n=-25

N-6=-12

n=-6

N+4=16

n=12

Step-by-step explanation:

5 0
3 years ago
True or False: Every number has a limited amount of factors, but an unlimited amount of multiples. ​
Irina18 [472]

Answer:

true

Step-by-step explanation:

3 0
3 years ago
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