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3 years ago

8. What will be the extension of a spring of K = 10 N/m, if the load is 75g?

Physics

1 answer:

ira [324]3 years ago

4 0

Answer:

<em>The extension of the spring is 7.35 cm</em>

Explanation:

<u>Spring Force </u>

When a spring is compressed or extended by a distance x, it exerts a force given by the Hooke's Law:

F=k.x

Where k is the constant of the spring.

A mass of m=75 gr is hung to the spring of constant k=10 N/m. The force applied to the spring is the weight of the mass:

W = mg

The mass must be converted to kg:

m = 75/1000 = 0.075 kg

W=0.075*9.8

W = 0.735 N

The extension of the spring can be computed by solving for x:

$\displaystyle x=\frac{0.735}{10}$

x=0.0735 m

x = 7.35 cm

The extension of the spring is 7.35 cm

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Answer:

q/m = 2177.4 C/kg

Explanation:

We are given;

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Also, deflection down is given by;

d_1 = ½at²

Now,we know that in electric field;

F = ma = qE

Thus, a = qE/m

So;

d_1 = ½ × (qE/m) × (D_1/v_o)²

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The deflection due to this is;

d_2 = V_y × t_2

Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)

d_2 = (qE/m) × (D_1•D_2/(v_o)²)

Total deflection down is;

d = d_1 + d_2

d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]

d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]

Making q/m the subject, we have;

q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]

We have;

E = 800 N/C

d = 1.25 cm = 0.0125 m

D_1 = 26.0 cm = 0.26 m

D_2 = 56 cm = 0.56 m

Thus;

q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]

q/m = 312500/143.52

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Answer:

t = 12.8 s

Explanation:

As we know that the deepest point of the canyon is 800 m

so here we will have the displacement of the rock in the canyon is same as that of depth of the canyon

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$\Delta y = v_y t + \frac{1}{2}at^2$

$800 = 0 + \frac{1}{2}(9.8) t^2$

$800 = 4.9 t^2$

now we have

$t^2 = \frac{800}{4.9}$

$t = \sqrt{163.6}$

$t = 12.8 s$

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4 years ago

A wave travels at a constant speed. how does the wavelength change if the frequency is reduced by a factor of 3? assume the spee

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Answer:

Explanation:

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2 years ago

Strzała o masie 20g tuz po wystrzale ma prędkość 50m/s. Oblicz pracę wykonana przez zawodniczkę. Ile wynosi energia potencjalna

garri49 [273]

Answer:

Explanation:

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3 years ago

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

4 0

3 years ago

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8. What will be the extension of a spring of K = 10 N/m, if the load is 75g?​

8. What will be the extension of a spring of K = 10 N/m, if the load is 75g?