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likoan [24]
3 years ago
13

How to find the angle marked

Physics
1 answer:
Sedbober [7]3 years ago
4 0
#6).  The tangent of an angle in a right triangle is

                                           (opposite side) / (adjacent side) .

The side opposite the marked angle is  4cm.
The side adjacent to the marked angle is  12cm.

The tangent of the marked angle is   (4cm / 12cm)  =  1/3 .

The marked angle is the angle whose tangent is 1/3 .
You can find that online, or with your calculator, slide rule, Curta, or book.


#7).  The cosine of an angle in a right triangle is

                                           (adjacent side) / (hypotenuse) .

The side adjacent to the marked angle is  15m.
The hypotenuse is  19m.

The cosine of the marked angle is   (15m / 19m)  =  about 0.78947...

The marked angle is the angle whose cosine is 0.78947... .
You can find that online, or with your calculator, slide rule, Curta, or book.
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It can be noted that the directions west and north are perpendicular to each other. hence we can use the right angle formula for calculating total distance in these directions. Also South is just negative north. This technique is called euclidean distance norm. It is the basis of Cartesian coordinate system.

Total distance = 3.22 N + 4.75 W +1.90  S

= 1.32 N + 4.75 W

D = \sqrt{N^2 + W^2}

D = 4.93 Km

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3 years ago
If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit
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Answer:

v = r\omega

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

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The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

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A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
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A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

k=463.28\ N/m

3 0
3 years ago
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