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Kipish [7]
4 years ago
7

The magnitude and direction of two forces acting on an object are 100100 ​pounds, S7878degrees°​E, and 5050 ​pounds, N5353degree

s°​E, respectively. Find the​ magnitude, to the nearest hundredth of a​ pound, and the direction​ angle, to the nearest tenth of a​ degree, of the resultant force.

Mathematics
1 answer:
amm18124 years ago
4 0

Answer:

  138.06 lb N86.1°E

Step-by-step explanation:

There are several ways you can work this. One of the most straightforward is to resolve each vector into its north and east components, add those, and then covert the result back to magnitude and direction.

A diagram can help immensely.

We note that S78°E is the same as E12°S. Since we're used to seeing the coordinate system with the +x axis aligned with east, it can be convenient to think of the first force as 100 lb at -12°.

The angle of the second force is N53°E, which can be expressed as E37°N. Then in x-y coordinates, this force is 50 lb at +37°.

The components of the sum are the sum of the components:

  R = F1 +F2 = (100cos(-12°) +50cos(37°), 100sin(-12°) +50sin(37°))

  = (137.747, 9.300)

The the magnitude of the resultant is computed using the Pythagorean theorem:

  ||R|| = √(137.747² +9.300²) ≈ 138.06

and the angle is computed using the arctangent function. Here, our diagram tells us the angle is in the first quadrant, so is positive (relative to +x, or East).

  ∠R = arctan(9.300/137.747) ≈ 3.9°

__

We want to express the answer in terms similar to the way the given forces are expressed, so we want the angle relative to north. The resultant is then described by ...

  R = 138.06 lb at N86.1°E

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Answer:

Step-by-step explanation:

Here you go mate

Step 1

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Step 2

-3[5-(-8+6)]  Simplify

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Hope this helps

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3 years ago
Whats the algebraic justification for this?
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3 0
3 years ago
A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar
Arisa [49]

Answer:

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

n=17 represent the sample selected

\alpha=0.1 represent the significance

s^2 =4.7^2 =22.09 represent the sample variance

\sigma^2_0 =3.5^2 =12.15 represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 12.15

Alternative hypothesis: \sigma^2 \neq 12.15

The statistic for this case is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

The degrees of freedom are:

df =n-1= 17-1=16

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0
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devlian [24]
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       Answer: The area of the pentagon ABCDE is 341.9 cm².
8 0
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