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3 years ago

Un resorte se monta horizontalmente con su extremo izquierdo fijo. Conectando una balanza de

Physics

1 answer:

zavuch27 [327]3 years ago

7 0

Translation of important question part (Google translation used)

Force of 6N causes a displacement of 0.03m. remove the balance and connect a body 0.5 kg to the end, pull it to move it 0.02 m, release it and see how it oscillates.

(a)Determine the spring constant.

(b)Calculate angular velocity, frequency, and oscillation period

Answer:

(a)K=200 N/m

(b) w= 20 rad/s f=3.2 Hz T=0.3125 s

Explanation:

(a)

From Hooke's law, we deduce that F=kx where F is applied force, k is spring constant and x is extension of the spring. Making k the subject of the formula then k=F/x and substituting F with 6 N and x with 0.03 m then k=6/0.03=200 N/m

(b)

Angular velocity, w is given by $w= \sqrt{\frac {k}{m}}$ where m is the mass and k is spring constant calculated in part a above. Substituting mass with 0.5 kg and k with 200 N/m then

$w= \sqrt{\frac {200}{0.5}}=20 rad/s$

We know that frequency, f is given by $f=\frac {w}{2\pi}$ and substituting 20 rad/s for w then

$f=\frac {20}{2\pi}=3.1830988618379067153776752674502872406891\approx 3.2 Hz$

Finally, oscillation period, T is usually the reciprocal of frequency hence T=1/f and substituting f with 3.2 Hz then T=1/3.2=0.3125 s

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3 years ago

A long straight wire carries current to the right of the page. A rectangular loop is positioned directly under the wire in the p

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Answer:

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Explanation:

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Given that,

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Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

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3 years ago

Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 25.0 m from each other. Bein

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Answer:

A) 199.78 J

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C) 4.2x10^7 m/s

D) 0.65 m

E) 1.13x10^-8 sec

D) 2.94x10^-9 sec

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mass of ball = 0.0580 kg

If smashed at v = 83.0 m/s, KE is

KE = 0.5mv^2

= 0.5 x 0.0580 x 83.0^2

= 199.78 J

B) if returned at v = 1.79×10^8 m/s, KE will be

KE = 0.5mv^2

= 0.5 x 0.0580 x (1.79×10^8)^2

= 9.292x10^14 J

C) during Einstein's return, velocity of rabbit relative to players is

Vr = 2.21×108 m/s

Rabbit's velocity relative to ball = 2.21×10^8 - 1.79×10^8

= 4.2x10^7 m/s

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= 1.13x10^-8 (1 - 2.21/3)

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3 years ago

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Answer:

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Pebble: $F = 29.44N$

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Pebble: $a =9.8m/s^{2}$

Explanation:

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m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

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$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

Case for the pebble $m = 3.0 Kg$ :

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

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