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3 years ago

15

A 50 meter chain with uniform density and a mass of 40 kilograms is hanging from the roof of a building under construction. How

much work (in Nm) is required to pull the chain up?

2 answers:

Brums [2.3K]3 years ago

8 0

Answer:

19,600 Nm

Step-by-step explanation:

WD = mgh

= 40 × 9.8 × 50

= 19600 Nm

sertanlavr [38]3 years ago

5 0

Answer:

W = 19600 N m

Step-by-step explanation:

W = force * distance

The force is the weight times gravity since it is hanging down

W = 40 kg* 9.8 m/s^2 * 50m

19600 kg m^2/s^2

A newton is kg m/ s^2

W = 19600 N m

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Please help. Thank you.

Marysya12 [62]

The answer would be 24pi inches.

You can find this by using the formula for a circumference and plugging in the value of the radius.

C = 2pi*r

C = 2pi * 12

C = 24pi

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3 years ago

Gretta is 1 1/2 meters tall. Which of the following is equivalent to 1 1/2 meters?

Furkat [3]

The answer would be one of the following in the photo below....if not do you have a picture you could show me of the possible answers?

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3 years ago

The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,

uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

$\frac{500-100}{100} = A}$ and $A=4$ .

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$ .

Finally, as we want the value of $t$ such that $P(t)=200$ , we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$ .

This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$ .

So, the population of rats will be 200 after 25 months.

6 0

3 years ago

Find the area of the following shape,

lord [1]

$\bold{ANSWER:}$
125.71 cm^2

$\bold{SOLUTION:}$

5 0

2 years ago

What would the amplitude be? How do I find it? Should I add 1.25+6.75 then divide by 2 ?

Elis [28]

Answer:

2.75

Step-by-step explanation:

The amplitude is half of the difference between the highest and lowest y-coordinates.

amplitude = 0.5|6.75 - 1.25| = 0.5(5.5) = 2.75

3 0

2 years ago