All categories

3 years ago

Part a

1 answer:

8 0

You need to find the acceleration once the rope starts acting.

For that, first you need the velocity, V, when Karen ends the 2.0 m free fall

V^2 = Vo^2 + 2gd = 0^2 + 2* 9.81m/s^2 * 2.0 m = 39.24 m^2 / s^2 => V = 6.26 m/s

From that point, the rope starts acting with a net force that produces a constant acceleration, modeled as per these values and equation:

Vo = 6.26m/s
Vf = 0

Vf^2 = Vo^2 + 2ad => a = [Vf^2 - Vo^2] / 2d = [0^2 - (6.26 m/s)2 ] /2(1m) = -19.62 m/s^2

That acceleration is due to the difference of the force applied by the rope - the weight of Karen:

Weight of Karen: mg

Weight of karen - Force of the rope = Net force = ma

mg - Force of the rope = ma

Force of the rope = mg - ma = m(g -a) = m [9.81m/s^2 - (-19.62m/s^2) ]= 29.43m

To express it as a multiple of her weight, divide it by her weight (mg = 9.81m) =>

29.43m / 9.81m = 3.0

Answer: 3.0 times.

You might be interested in

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

A jacket at Academy is $60. You have a coupon for 20% off. What is the new price?

svetlana [45]

Answer:

$48

Step-by-step explanation:

7 0

3 years ago

An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 8 sees

zheka24 [161]

Answer:

Probability = 0.12025

Step-by-step explanation:

P (Am) = 1/50 = 0.02 {Magazine ad}

P (At) = 1/8 = 0.125 {Television ad}

P (Am ∩ At ) = 1/100 = 0.01 {Both ads}

P (Am U At) = P (Am) + P (At) - P (Am ∩ At )

= 0.02 + 0.125 - 0.01

P (Am U At) = 0.135 {Person sees either ad}

P (Am' ∩ At') = 1 - P (Am U At)

P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}

Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)

P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1

P (P) = (0.25) (0.135) + (0.1) (0.865)

= 0.03375 + 0.0865

0.12025

7 0

2 years ago

I need help on this it’s about trig identities

Cerrena [4.2K]

Answer:

The answer to your question is:

Step-by-step explanation:

1.-

$\frac{1 + sin\alpha }{cos\alpha } + \frac{cos\alpha }{1 + sin\alpha } = 2 sec\alpha$

$\frac{(1 + sin\alpha)^{2} + cos^{2} \alpha }{cos\alpha (1 + sin\alpha) }$

$\frac{1 + 2sin\alpha + sin^{2} \alpha+ cos^{2} \alpha }{cos\alpha + sin\alphacos\alpha }$

$\frac{2 + 2sin\alpha }{cos\alpha+ sin\alpha cos\alpha }$

$\frac{2(1 + sin\alpha) }{cos\alpha(1 + sin\alpha ) }$

$\frac{2}{cos\alpha }$

2sec $\alpha$

2.-

sec²x - tanxsecx

$\frac{1}{cos^{2}x } - \frac{sinx}{cosx} \frac{1}{cosx}$

$\frac{1}{cos^{2} x} - \frac{sinx}{cos^{2}x}$

$\frac{1 - sinx}{cos^{2}x }$

$\frac{1 - sinx}{1 - sin^{2}x }$

$\frac{1 - sinx}{(1 - sinx)(1 + sinx)}$

$\frac{1}{1 + sinx}$

3 0

3 years ago

Which two values of x are the roots of the polynomial below? <br> x^2+5x+11

In-s [12.5K]

Answer:

x = -5/2 + i√19 and x = -5/2 - i√19

Step-by-step explanation:

Next time, please share the possible answer choices.

Here we can actually find the roots, using the quadratic formula or some other approach.

a = 1, b = 5 and c = 11. Then the discriminant is b^2-4ac, or 5^2-4(1)(11). Since the discriminant is negative, the roots are complex. The discriminant value is 25-44, or -19.

Thus, the roots of the given poly are:

-5 plus or minus i√19

x = -----------------------------------

2(1)

or x = -5/2 + i√19 and x = -5/2 - i√19

5 0

3 years ago

Read 2 more answers