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maxonik [38]
3 years ago
14

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.
Physics
1 answer:
levacccp [35]3 years ago
5 0

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

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<span>China</span>

7 0
3 years ago
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A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant lin
Lapatulllka [165]

Answer:

Kr = 0.7618K

Explanation:

Suppose that the object's velocity is V, then his kinetic energy is:

K = \frac{mv^{2} }{2}

K = \frac{(16)v^{2} }{2}

K = 8v^{2}

The rotational kinetic energy is

Kr = \frac{Iw^{2} }{2}

           where I: The moment of inertia

                      ω: angular velocity

Kr =\frac{(0.59)w^{2} }{2}

Kr = 0.295w^{2}

How the movement is without slipping, then  

ω = \frac{v}{r}

ω = \frac{v}{0.22}

Thus

Kr = \frac{0.295v^{2} }{0.22^{2} }

Kr = 6.095v^{2}

8v^{2}  ---->  1

6.095v^{2}----->?

Kr = 0.7618K

4 0
4 years ago
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Answer:

d

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3 0
3 years ago
Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has
k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

8 0
3 years ago
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slava [35]

Answer:

What inferences can you make about the melting points of the different substances and the motion of their particles, based on the data? (ignore that needed it here so i could see it better.)

Explanation:

Butter has a lower melting point than the cheese and the wax. The motion of the cheese were a little separated while the butter articles have more space in between. The wax had the closest particles.

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5 0
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