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liq [111]
3 years ago
9

A 5.8kg box is on a frictionless 45∘ slope and is connected via a massless string over a massless, frictionless pulley to a hang

ing 2.1kg weight.What is the tension in the string if the 5.8kg box is held in place, so that it cannot move?What is the tension in the string once the box begins to move?
Mathematics
1 answer:
nexus9112 [7]3 years ago
5 0

Answer:

a) T = 20.601 N

b) T = 9.4585 N

Step-by-step explanation:

The tension in the string if the 5.8 kg box is held in place, so that it cannot move can be obtained as follows

For m = 2.1 Kg

∑ Fy = 0

T - W = 0  ⇒  T = W = m*g

⇒  T = 2.1 Kg*9.81 m/s²

⇒  T = 20.601 N  (↑)

The tension in the string once the box begins to move can be obtained as follows

For M = 5.8 Kg

∑ Fx' = M*a

where x' is the direction of the slope

then

∑ Fx' = M*a   ⇒  <em>T - M*g*Sin ∅ = M*a    (I)</em>

For m = 2.1 Kg

∑ Fy = m*a

⇒  T - W = m*a

⇒  <em>T - m*g = m*a    (II)</em>

If we solve the system of equations that comprises <em>I</em> and <em>II</em> we will know T:

T = M*m*g*(Sin ∅ -1) / (m - M)

⇒  T = (5.8 Kg)(2.1 Kg)(9.81 m/s²)(Sin 45º - 1) / (2.1 Kg - 5.8 Kg)

⇒  T = 9.4585 N

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At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

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Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=3278.222

The sample deviation calculated s=97.054

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Now we have everything in order to replace into formula (1):

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

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c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

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