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son4ous [18]
2 years ago
8

A cave explorer travels 3.0 m eastward, then 2.5 m northward, and finally 15.0 m westward. use the graphical method to find the

magnitude of the net displacement. scale: ________________ magnitude: ___________________ direction: ____________________ can someone help me please i have an exammm tomorrow and i dont know who to solve it ...

Physics
2 answers:
astraxan [27]2 years ago
8 0

Answer:

12.26 m, north-west

Explanation:

From the figure the displacement can be calculated by the Pythagorean law. In this the triangle ABC is the right angle triangle with hypotenous h, base 12 m, and height 2.5 m.

Therefore,

h^{2}=12^{2}+2.5^{2}\\   h=\sqrt{150.65} \\h=12.26m

Therefore the net displacement is 12.26 m and the displacement we know it is vector quantity and it is the shortest distance between the two points.

Therefore the displacement is from point B to A. Then the net displacement is the direction in the north-west direction with a magnitude of 12.26 m.

Dmitrij [34]2 years ago
7 0
3m east and 15m west evens out to 12m west. That's one side of a triangle. The second side is 2.5m north.
The third side is your displacement,
use Pythagorean's Theorem to determine this.
 a^2+b^2=c^2, that works out to 12^2+2.5^2=c^2.....144+6.25=c^2
<span>150.25=c^2
and
finally
c=12.26,
or 12m</span>
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creativ13 [48]
First of all, we can find the mass of the person, since we know his weight W:
W=mg=0.70 kN=700 N
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m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
\sum F =  ma
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2 years ago
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VladimirAG [237]

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A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

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The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

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C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

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8 0
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Inessa [10]
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6 0
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PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

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   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0
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