The magnitude and direction of two forces acting on an object are 100100 ​pounds, S7878degrees°​E, and 5050 ​pounds, N5353degrees°​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.

Answers

Answer 1
Answer:

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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Answers

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Select all sets that contain the number 0.484848.A. Integer
B. Irrational
C. Natural
D. Rational
E. Whole

Select all that apply

Answers


0.484848  . . . . .

A. Integer . . . no, it's not
B. Irrational . . . no, it's not
C. Natural. . . no, it's not
D. Rational . . . Yes !  It is !
E. Whole
. . . no, it's not


Why can't you factor 2cosx^2+sinx-1=0 ?

Answers

2cos^2x+sinx-1=0\n\n2(1-sin^2x)+sinx-1=0\n\n2-2sin^2x+sinx-1=0\n\n-2sin^2x+sinx+1=0\n\n-2sin^2x+2sinx-sinx+1=0\n\n-2sinx(sinx-1)-1(sinx-1)=0\n\n(sinx-1)(-2sinx-1)=0\iff sinx-1=0\ or\ -2sinx-1=0\n\nsinx=1\ or\ -2sinx=1\n\nsinx=1\ or\ sinx=-(1)/(2)\n\nx=(\pi)/(2)+2k\pi\ or\ x=-(\pi)/(6)+2k\pi\ or\ x=(7\pi)/(6)+2k\pi\ where\ k\in\mathbb{Z}
2cosx^2+sinx-1=2(1-sin^2x)+snx-1=\n\n=2(1-sinx)(1+sinx)-(1-sinx)=(1-sinx)[2(1+sinx)-1]=\n\n=(1-sinx)(2+2sinx-1)=(1-sinx)(1+2sinx)\n\n2cosx^2+sinx-1=0\ \ \ \Leftrightarrow\ \ \ (1-sinx)(1+2sinx)=0\n\n1-sinx=0\ \ \ \ \ or\ \ \ \ \ 1+2sinx=0\n\n1)\ \ \ 1-sinx=0\ \ \ \Rightarrow\ \ \ sinx=1\ \ \ \Rightarrow\ \ \ x= ( \pi )/(2) +2k \pi ,\ \ \ k\in I\n\n

2)\ \ \ 1+2sinx=0\ \ \ \ \ \ \Rightarrow\ \ \ sinx=- (1)/(2)\n\n \Rightarrow\ \ \ x_1=( \pi + ( \pi )/(6) )+2k \pi ,\ \ \ \ \ \ x_2=( - ( \pi )/(6) )+2k \pi,\ \ \ \ \ \ k\in I\n\n.\ \ \ \ \ \ x_1=(7 \pi )/(6) +2k \pi ,\ \ \ \ \ \ \ \ \ \ \ \ x_2=-( \pi )/(6) +2k \pi,\ \ \ \ \ \ \ \ \ k\in I\n\nAns.\ x=-( \pi )/(6) +2k \pi\ \ \ or\ \ \ x= ( \pi )/(2) +2k \pi\ \ \ or\ \ \ x=(7 \pi )/(6) +2k \pi,\ \ \ k\in I

Solve for
a. 7a-2b = 5a b

Answers

For this case we have the following expression:

From here, we must clear the value of a.

We then have the following steps:

Place the terms that depend on a on the same side of the equation:

Do common factor "a":

Clear the value of "a" by dividing the factor within the parenthesis:

Answer:

The clear expression for "a" is given by:

Answer:  The required value of a is (2b)/(7-5b).

Step-by-step explanation:  We are given to solve the following equation for the value of a:

7a-2b=5ab~~~~~~~~~~~~~~~~~~~~~(i)

Since there are two unknowns and only one equation , so the value of a will definitely contain the value of b.

The solution of equation (i) for a is as follows:

7a-2b=5ab\n\n\Rightarrow 7a-5ab=2b\n\n\Rightarrow a(7-5b)=2b\n\n\Rightarrow a=(2b)/(7-5b).

Thus, the required value of a is (2b)/(7-5b).