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My name is Ann [436]

2 years ago

11

An elevator weighing 15,000 newtons is raised one story in 50 seconds. If the distance between stories is 3.0 meters, how much p

ower is required of the motor? 750,000 W 45,000 W 5,000 W 900 W

2 answers:

Pavel [41]2 years ago

8 0

Power
= Work done / Time
= 15000 N * 3 m / 50 s
= 45000/50 N-m/s
= 900 W

It may sound like an extremely low value, but 3.0 / 50 s is also a very slow speed.

Trava [24]2 years ago

8 0

The answer is:

900 Watts

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After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the maximum

Svetlanka [38]

Answer: 40000

Step-by-step explanation:

The formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$ , where p is the prior estimate of the population proportion.

Here we can see that the sample size is inversely proportion withe square of margin of error.

i.e. $n\ \alpha\ \dfrac{1}{E^2}$

By the equation inverse variation, we have

$n_1E_1^2=n_2E_2^2$

Given : $E_1=0.05$ $n_1=1000$

$E_2=0.025$

Then, we have

$(1000)(0.05)^2=n_2(0.025)^2\\\\\Rightarrow\ 2.5=0.000625n_2\\\\\Rightarrow\ n_2=\dfrac{2.5}{0.000625}=4000$

Hence, the sample size will now have to be 4000.

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2 years ago

The product of five rational numbers is positive. At most, of these rational numbers can be negative

Ivanshal [37]

Answer: A. 1

Step-by-step explanation:

sorry if im wrong

5 0

2 years ago

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

4 0

2 years ago

In a survey, a group of students were asked to name their favorite elective class. There were 15 students who chose “other." How

lyudmila [28]

Answer:

50 % of them participated

7 0

2 years ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

$\large t_{0.05}=2.365$

Now we can compute our confidence interval

$\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546$

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

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2 years ago