964 nf Homework 2 Print Assignment Copy of Homework 1 faise Chapter 01, Problem

Question

964 nf Homework 2 Print Assignment Copy of Homework 1 faise Chapter 01, Problem 36 Soccer player # 1 is 8.08 m from the goal, asthe figure shows. If she kicks the ball directly into the net, the ball has a displacement labeled A. If, on the other hand, she first kicks it to player#, who then kicks it into the ne and Av. What are the magnitudes of (a) Ax, and (b) áy t, the ball undergoes two successive displacements, A (a) Number 04Units (b) Number947Units " significant digits are disabled; the tolerance is +/-5% , significant digits are disabled; the tolerance is +/-5% Chapter 01, Problem 52 GO Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the location of the first team as 35 km away, 24° north of west, and the second team as 31 km away, 31° east of north. When the first team uses its GPS the position of the second team, what does it give for the second team's (a) distance from them and (b) direction, measured from due east? (o) Number 4 1, (b) Number 1+. Unts Kn Units significant digits are disabled; the tolerance is +/-590 7, significant digits are disabled; the tolerance is +/-5% "3-significant digits are disabled; the tolerance is +/-5% 4, significant digits are disabled; the tolerance is +/-5% significant digits are disabled, the tolerance is +/-5% "6-significant digits are disabled; the tolerance is +/-5% Chapter 02, Problem 02 GO

Explanation / Answer

1.

Suppose given that Vector is R and it makes angle X with +x-axis, then it's components are given by:

Rx = R*cos X

Ry = R*sin X

Using above rule We need to find components of above vector

Given that A = 8.08 m at the angle of 30 deg East of South

A = 8.08 m, at 60 deg South of East

Now

Ax = 8.08*cos 60 deg = 4.04 m

|Ax| = 4.04 m

Ay = -8.08*sin 60 deg = -6.997 m

|Ay| = 7.00 m (We need final answer in 3 significant digits)

2.

Again Using above rule, Position of team 1 is

P1 = 35 km, at 24 deg North of West

P2 = 31 km, at 31 deg east of north = 31 km, at 59 deg North of East

Now we need distance between both tems, which will be given by:

P = Px i + Py j

|P| = sqrt (Px^2 + Py^2)

Direction = arctan (Py/Px)

Now

P1x = -P1*cos A1 = -35*cos 24 deg = -31.97 km

P1y = P1*sin A1 = 35*sin 24 deg = 14.23 km

P2x = P2*cos A2 = 31*cos 59 deg = 15.97 km

P2y = P2*sin A2 = 31*sin 59 deg = 26.57 km

Now

P = (P1x + P2x) i + (P1y + P2y) j

P = (-31.97 + 15.97) i + (14.23 + 26.57) j

P = -16 i + 40.8 j

|P| = sqrt ((-16)^2 + 40.8^2) = 43.82 km

Direction = arctan (40.8/16) = 65.59 deg from -ve x-axis (west) = 180 - 65.59

Direction = 114.41 deg due east

Please Upvote.

Comment below if you have any doubt.

Related Questions

Navigate

• Browse (All)
• Browse 9
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.