This question that it pictured

A quelle distance le soletl se trouve-t-il de la terre?

Mnenie [13.5K]

The explanation for the following answer is explained below.

Explanation:

The sun is at an average distance of about 93,000,000 miles(150 million kilometers) away from the earth.It is so far away that light from the Sun,travelling at a speed of 186,000 miles (300,000 kilometers) per second, takes about 8 minutes to reach the earth.Earth does not travel around the Sun in a perfect circle.Instead its orbit is elliptical,like a stretched circle,with the sun just off the center of the orbit. At its closest,the Sun is 91.4 million miles (147.1 million kilometers ) away us.At its farthest ,the Sun is 94.5 million miles (152.1 million km) away.The Earth is closest to the Sun during winter in the northern hemisphere

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 spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ

emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0

3 years ago

How to measure the volume of a baseball bat ( need answers ASAP )

vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water. But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly. Some water will run over the top and out of the tub. That's OK ... it's exactly what you want. If NO water runs over the top, pour some more in, until it runs out and then stops. You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan. The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string. Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it. The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub. Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub. Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it. Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions. Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces. But you can find all those conversion factors with a search on Floogle.

8 0

3 years ago

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$ .

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$

8 0

2 years ago