All categories

4 years ago

For a spring oscillating in simple harmonic motion, at what point will the velocity of

Physics

1 answer:

Crazy boy [7]4 years ago

6 0

Answer:

equilibrium position.

Explanation:

In simple harmonic motion , velocity v(t) is given by,

v(t) = -ω A sin(ωt + φ)

where

ω = angular velocity of the corresponding circular motion

A = amplitude

t = time

φ = the initial angle of the corresponding circular motion when the motion begin.

v (t) get maximized when sin value is maximized , i.e. sin $\alpha$ =1

The particle has maximum speed when it passes through the equilibrium position.

You might be interested in

A positive charge is placed in an electric field that points west. What direction is the force on the positive particle? I think

Kisachek [45]

Right. You are true. The direction of the electric field is defined to be
the direction of the force on a small positive charge placed in the field.

3 0

4 years ago

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

3 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es:

$\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}$

$\Delta S = 214.235 \,\frac{J}{K}$

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0

3 years ago

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

charle [14.2K]

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

4 0

3 years ago