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Sloan [31]
3 years ago
8

The formula represents the kinetic energy K of an object with mass m and velocity v. K=12mv2 Solve for v. v=±12Km2 v=±2K√m v=±2K

m‾‾‾‾‾√ v=±2Km√m
Mathematics
2 answers:
Elis [28]3 years ago
6 0
<span>v = ±2K√m
Hope this helps </span><span />
Kazeer [188]3 years ago
5 0

Answer:

Option 3rd is correct

v = \pm \sqrt{\frac{2K}{m}}

Step-by-step explanation:

Given that:

The formula represents the kinetic energy K of an object with mass m and velocity v is given by:

K = \frac{1}{2}mv^2          ....[1]

Solve for v:

Multiply both sides by 2  in [1] we have;

2K = mv^2

Divide both sides by m we get;

\frac{2K}{m} = v^2

or

v^2 = \frac{2K}{m}

⇒v = \pm \sqrt{\frac{2K}{m}}

Therefore, the value of v is, v = \pm \sqrt{\frac{2K}{m}}

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Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.
garik1379 [7]

Answer:

The 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

Step-by-step explanation:

Previous concepts  

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_1 =9.2 represent the sample mean 1

\bar X_2 =8.8 represent the sample mean 2

n1=27 represent the sample 1 size  

n2=30 represent the sample 2 size  

\sigma_1 =0.3 population standard deviation for sample 1

\sigma_2 =0.1 population standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =9.2-8.8=0.4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96  

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

0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281  

0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519  

So on this case the 95% confidence interval would be given by 0.281 \leq \mu_1 -\mu_2 \leq 0.529  

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4 years ago
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Answer:

easy!

Step-by-step explanation:

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debby collected eight times as many bags of cans as her friend .if her friend collected two thirds of a bag. how many bags did d
sergij07 [2.7K]

Answer:

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Step-by-step explanation:

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Train A and train B leave the station at 2 pm. The graph below shows the distance covered by the two trains. Compare the two spe
jarptica [38.1K]

Both trains travel for 7 time units.

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\dfrac{385}{7}=55

Train B travels 577.5 distance units, so its speed is

\dfrac{577.5}{7}=82.5

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