Ask question

Ask question

All categories

2 years ago

13

Using an ideal machine a worker exerts an effort force of 8.00 N to life a 2.95 kg weight a distance of 6:00 am over what distan

ce does the worker apply his force?

1 answer:

cestrela7 [59]2 years ago

3 0

Answer:

29.5 ×6= 8×d

d = 29.5×6/8 = 22.12 m

You might be interested in

What type of function can be used to model the scenario?

11111nata11111 [884]

Answer:

linear

Step-by-step explanation:

7 0

2 years ago

Please explain how to do and set up

In-s [12.5K]

A

using the Cosine rule in ΔSTU

let t = SU, s = TU and u = ST, then

t² = u² + s² - (2us cos T )

substitute the appropriate values into the formula

t² = 5² + 9² - (2 × 5 × 9 × cos68° )

= 25 + 81 - 90cos68°

= 106 - 33.71 = 72.29

⇒ t = $\sqrt{72.29}$ ≈ 8.5 in → A

6 0

3 years ago

I need help with one question. <br> anyone pls?

oksano4ka [1.4K]

Answer:

(-3,4)

Step-by-step explanation:

x + 2y = 5

-3 + (2*4) = 5

-3 + 8 = 5

5 = 5

2x + 3y = 6

(2*-3) + (3*4)=6

-6 + 12 = 6

6 = 6

4 0

2 years ago

Read 2 more answers

Find the distance between the points A and B given below.

GalinKa [24]

Answer:

use formula

Step-by-step explanation:

distance between the points A and B=

√((x2-x1)²+(y2-y1)²)

5 0

2 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

2 years ago