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Tamiku [17]
3 years ago
15

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and directio

n of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components

Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°

Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.

Their resultant is gotten below

Magnitude of resultant = 3.25 - 1.50 = 1.75km

Direction of resultant = north (direction of the bigger vector)

2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.

R = √ vx² + vy²

R = √ 2.20² + 1.75²

R = √ 7.9025

R= 2.1811km.

θ = tan^-1 (vy/vx)

θ = tan ^-1 (1.25/2.20)

θ = tan ^-1 (0.5618)

θ = 29.6°.

The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ

Thus the direction of the vector is 180 + 29.60 = 209.6°

From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.

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Explanation:

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          P = V_s i

          V_s = P / i

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           \frac{V_p}{N_p} = \frac{V_s}{N_s}

           \frac{N_s}{N_p} = \frac{V_s}{V_p}

           Ns / Np = 13,818 10³ /120

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
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Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

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4 years ago
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Answer:

crystallography.......

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