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2 years ago

What is the velocity of an object dropped from a height of 180 m when it hits the ground?

2 answers:

7 0

You can use this formula: <span>vf^2=0+2(9.8)(M)
solve it out like so:
vf^2=0+2(9.8)(180)
vf^2=2(1764)
vf^2=3528
now, square both sides to cancel out the power of 2. you are left with:
about 59.4 m/s</span>

s2008m [1.1K]2 years ago

4 0

V^2 = initial velocity^2 + 2* acceleration * distance

v^2 = (0m/s) + 2* 9.8 m/sec^2 * 180 m

v^2 = 3528 m^2 / sec^2

v = 59.397 m/sec

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Question 3 of 8. Step 1 of 1

natali 33 [55]

308 gallons of 7% and 77 gallons of 2% are needed to obtain the desired 385 gallons.

<h3><u>Combination</u></h3>

Since a dairy needs 385 gallons of milk containing 6% butterfat, to determine how many gallons each of milk containing 7% butterfat and milk containing 2% butterfat must be used to obtain the desired 385 gallons, the following calculation must be performed:

385 x 0.06 = 23.1
300 x 0.07 + 85 x 0.02 = 22.7
310 x 0.07 + 75 x 0.02 = 23.2
308 x 0.07 + 77 x 0.02 = 23.1

Therefore, 308 gallons of 7% and 77 gallons of 2% are needed to obtain the desired 385 gallons.

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1 year ago

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san4es73 [151]

Answer:

Step-by-step explanation

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2 years ago

Medical scientists study the effect of acute infection on tissue-specific immunity. In a collection of experiments under the sam

Serjik [45]

Answer:

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. This means that $n = 75$ and $\pi = \frac{44}{75} = 0.586$ .

Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 - 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.5291$

The upper limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 + 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.6429$

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

7 0

2 years ago

Can somebody please help me? I will mark brainiest if i can!<br> qr + r; use q = 3, and r = -6

Sveta_85 [38]

The answer is -12. R times -6 is -18 and -18 plus -6 is -12

5 0

2 years ago

Read 2 more answers

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

<u>Step-by-step explanation:</u>

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

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2 years ago