What is the smallest unit that makes up matter

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

6) On Earth, the gravity acceleration is $g=9.81 m/s^2$ . The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
 $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$

5 0

3 years ago

A machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars comple

babunello [35]

Answer:

a) Null hypothesis: $\mu \geq 16$

Alternative hypothesis : $\mu$

b) $t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77$

$p_v =P(t_{(7)}$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

c) We can conclude that the mean is significantly less than 16 ounces at 10% of significance. so then the consumer suspect is correct.

Explanation:

Data given and notation

$\bar X=15.6$ represent the mean for the sample

$s=0.3$ represent the sample standard deviation

$n=8$ sample size

$\mu_o =16$ represent the value that we want to test

$\alpha=0.1$ 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.

Is a one tailed left test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \geq 16$

Alternative hypothesis : $\mu$

The degrees of freedom on this case are:

$df=n-1=8-1=7$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77$

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:

$p_v =P(t_{(7)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can conclude that the mean is significantly less than 16 ounces at 10% of significance. so then the consumer suspect is correct.

3 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was projected upward from the surface, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)