A 2.0 mass has j of kinetic energy <br> What is the speed?

Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physicallyrealistic. T

nata0808 [166]

Answer:

a. No, typical small pickup truck springs are not large enough to compress 0.61{\rm m}.

Explanation:

8 0

3 years ago

Help meeeeeeeeeeeee ill mark brainlist to people

kvasek [131]

Answer: question 1 , would be one question 2 , would be 1 joule and number three would be number one and number four would be , power and last one would be, number two

Explanation: sorry if its wrong

5 0

3 years ago

Hannah is very good at picking up differences in other people's mood and behavior. She has been rewarded for her ability to work

Elanso [62]

It's C. Most likely

5 0

4 years ago

Read 2 more answers

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 4.79 g coins stacked over the 28.8 cm mark, th

masha68 [24]

To solve the exercise, the key concept to be addressed is the Mass Center.

The center of mass of an object is measured as,

$X_{cm} = \frac{\sum m_ix_i}{\sum m_i}$

$X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

Our values are given by,

$x_1 = 50cm$

$m_1 = ?$

$x_2 =28.8cm$

$m_2 = 4.79g$

$x_3 = 28.8cm$

$m_3 = 4.79g$

$X_{cm} = 38.4cm$

Replacing the values in our previous equation we have,

$X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$38.4 = \frac{m_1(50)+2(28.8*4.79)}{m_1+2*4.79}$

$38.4(m_1+2*4.79)= m_1(50)+2(28.8*4.79)$

$38.4m_1 +367.872 = 50m_1+275.904$

$11.6m_1 = 91.968$

$m_1 = 7.928g$

Therefore the mass of the meter stick is 7.928g

7 0

3 years ago

For the population, scores on the test are normally distributed with μ = 70 and σ = 15. The sample of n = 25 students had a mean

AysviL [449]

Answer:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

$p_v =2*P(t_{24}>1.67)=0.108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

Explanation:

1) Data given and notation

$\bar X=75$ represent the sample mean

$\sigma=15$ represent the standard deviation for the population

$n=25$ sample size

$\mu_o =70$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is different from 70, the system of hypothesis would be:

Null hypothesis: $\mu = 70$

Alternative hypothesis: $\mu \neq 70$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{24}>1.67)=0.108$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

3 0

4 years ago

A 2.0 mass has j of kinetic energy What is the speed?