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3 years ago

How would you do this last part?

Mathematics

1 answer:

alina1380 [7]3 years ago

8 0

Answer:

broken heart thats the last one lol

Step-by-step explanation:

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A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Work done (Wd) =?

<h3>How to determine the spring constant</h3>

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>

Spring constant (K) = 5 pound/foot
Extention (e) = 0.7 feet
Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

5 0

1 year ago

What is the answer to the q

Anton [14]

Answer:

yes

Step-by-step explanation:

the explanation is that a rhombus is a parallelogram in which all sides are equal. Their diagonals bisect each other at right angles.

6 0

3 years ago

Suppose that y = k * (x - 1/3) ^ 2 is a parabola in the xy -plane that passes through the point (2/3, 1) :Find k and the length

Dennis_Churaev [7]

Answer:

k = 9

length of chord = 2/3

Step-by-step explanation:

Equation of parabola: $y=k (x-\frac13)^2$

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If the curve passes through point $(\frac23 ,1)$ , this means that when $x=\dfrac23$ , $y = 1$

Substitute these values into the equation and solve for $k$ :

$\implies 1=k \left(\dfrac23-\dfrac13\right)^2$

$\implies 1=k \left(\dfrac13 \right)^2$

Apply the exponent rule $\left(\dfrac{a}{b} \right)^c=\dfrac{a^c}{b^c}$ :

$\implies 1=k \left(\dfrac{1^2}{3^2} \right)$

$\implies 1=\dfrac{1}{9}k$

$\implies k=9$

The chord of a parabola is a line segment whose endpoints are points on the parabola.

We are told that one end of the chord is at $(\frac23 ,1)$ and that the chord is horizontal. Therefore, the y-coordinate of the other end of the chord will also be 1. Substitute y = 1 into the equation for the parabola and solve for x:

$\implies 1=9 \left(x-\dfrac13 \right)^2$

$\implies \dfrac19 = \left(x-\dfrac13 \right)^2$

$\implies \sqrt{\dfrac19} = x-\dfrac13$

$\implies \pm \dfrac13 = x-\dfrac13$

$\implies x=\dfrac23, x=0$

Therefore, the endpoints of the horizontal chord are: (0, 1) and (2/3, 1)

To calculate the length of the chord, find the difference between the x-coordinates:

$\implies \dfrac23-0=\dfrac23$

**Please see attached diagram for drawn graph. Chord is in red**

6 0

2 years ago

A use bike costs $75 a new one costs $300. how many times more expensive is a new bike compare with a use bime

Alisiya [41]

Answer:

jvjdebsjd bjiweibqewasje

Step-by-step expla

3 0

3 years ago

Read 2 more answers

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers