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motikmotik
2 years ago
9

A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​

Physics
1 answer:
Schach [20]2 years ago
5 0

Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

<h3>a = 0.04 m s²</h3>
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What does the term “retroperitoneal” mean in the terms of the location of the kidneys ?
lakkis [162]

Answer:

"Retroperitoneal" refers to the back of the peritoneum, the membrane that lines the anatomical space in the abdominal cavity. Kidney stones may cause pain to the organs within the retroperitoneal space. A diagram of the aorta, a retroperitoneal structure.

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5 0
3 years ago
A 20 ohm lamp and a 5 ohm lamp are connected in series and placed across a potential difference of 50 V.
iogann1982 [59]

Hi there!

1.

Since the two resistors are in series, we can simply add:
R_T = R_1 + R_2 + ... R_n

R_T = 20 + 5 = \boxed{25 \Omega}

2.

In series, the potential difference of each resistor (lamp) ADDS UP. We can begin by finding the current through the circuit using Ohm's law:
V = IR\\\\I = \frac{V}{R_T}

Plug in the values:
I = \frac{50}{25} = 2 A

Now,

we can use Ohm's law to find the individual voltage for each lamp.

20 Ohm lamp:
V = 2 * 20 = \boxed{40 V}

5 Ohm lamp:
V = 2 * 5 = \boxed{10 V}

3.

To solve, we can use the power equation.

P (\text{Watts})= IV

Plug in the values for each.

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5 Ohm lamp:
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7 0
1 year ago
Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum
Ronch [10]

Answer:

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Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ_{95%} = x_±(t*s)/sqrt(n)

where:

μ_{95%} = = is the 95% confidence interval estimate

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n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ_{95%} = 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

4 0
3 years ago
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5 0
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