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3 years ago

15

The plain flight covers 400 miles in the first hour. As you approach the landing site, it takes you an hour to cover the last 20

0 miles. This a example of
Positive acceleration.

Velocity.

Instantaneous speed.

Negative acceleration.

1 answer:

tatyana61 [14]3 years ago

3 0

that is an example of negative acceleration because it is slowing down

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the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.

asambeis [7]

$The\ area\ of\ a\ circle\ = \pi r^2 \\ 725\ cm^2 = 3.14r^2 \\
230.57 cm^2 = r^2 \\ \sqrt{230.57\ cm^2} = r \\ \boxed{r = 15.18\ cm}$

7 0

3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the

Gnesinka [82]

Answer:

F = 233.52 N, θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

= 80.5 N

Jill

cos 45 = F_{2x} / F₂2

sin 45 = F_{2y} / F₂2

F_{2x} = F₂ cos 45

F_{2y} = F₂ sin 45

F_{2x} = 81.7 cos 45 = 57.77 N

F_{2y} = 81.7 sin 45 = 57.77 N

Jane

cos (270 + 45) = F_{3x} / F₃3

sin 315 = F_{3y} / F₃

F_{3x} = 131 cos 315 = 92.63 N

F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

F_{x} = F_{1x} + F_{2x} + F_{3x}

F_{x} = 80.5 +57.77 + 92.63

F_{x} = 230.9 N

Axis y

F_{y} = F_{1y} + F_{2y} + F_{3y}

F_{y} = 0 + 57.77 -92.63

F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

F = √(Fₓ² + F_{y}²

F = √(230.9² + 34.86²)

F = 233.52 N

let's use trigonometry for the angle

tan θ = $\frac{F_y}{F_x} }$

θ = tan⁻¹ (\frac{F_y}{F_x} })

θ = tan⁻¹ (-34.86 / 230.9)

θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

θ' = 360 -θ

θ‘= 360- 8.59

θ' = 351.41º

5 0

3 years ago

Stephanie lives in California,and her cousin Thomas lives in France. which coordinate system would be most useful if the two wan

qwelly [4]

Answer:

Equatorial

Explanation:

Your welcomee

8 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

4 years ago

(blank) is the thermal energy that flows from one substance to another when the substances differ in temperature

Oduvanchick [21]

The answer is heat I think

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3 years ago

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