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3 years ago

Ayuda porfavor! y que sea buena respuesta..

Physics

1 answer:

Sav [38]3 years ago

8 0

Answer:

Dencidad = 600 kg/m³

Explanation:

Dados los siguientes datos;

Lado del cubo = 10 cm a metros = 10/100 = 0,1 m

Masa = 0,6 kg

Para encontrar la densidad;

En primer lugar, determinaríamos el volumen del cubo usando la fórmula;

Volumen del cubo = lados³

Volumen del cubo = 0.1³

Volumen del cubo = 0.001 m³

Ahora, podemos encontrar la densidad usando la fórmula;

Dencidad = masa/volumen

Dencidad = 0,6/0,001

Dencidad = 600 kg/m³

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An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from

uranmaximum [27]

Answer:

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Explanation:

From Hooke's law, we know that ;

F = kx

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So, Force = mg = 1.95 × 9.81

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Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

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mgL + ½k(x - L)² = ½kx² + ½mv²

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½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

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An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

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(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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seraphim [82]

Try this option, the answers are marked with colour.

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