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3 years ago

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A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

kg object be placed so as to experience a net force of zero from the other two objects?

1 answer:

djverab [1.8K]3 years ago

6 0

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

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balls a, with a mass of 20 kg, is moving to the right at 20 m/s. At what velocity should Ball B, with a mass of 40 kg, move so t

Leya [2.2K]

In that case, their momentum must be equal.
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10

In short, Your Answer would be: 10 m/s

Hope this helps!

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3 years ago

A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?

adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

$\sum F = 0$

$F_b -W = 0$

$F_b = W$

$F_b = mg$

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

$\rho_w V_{displaced} g = mg$

Remember the expression for which you can determine the relationship between mass, volume and density, in which

$\rho = \frac{m}{V} \rightarrow m = V\rho$

In this case the density would be that of the object, replacing

$\rho_w V_{displaced} g = V\rho g$

Since the displaced volume of water is 0.429 we will have to

$\rho_w (0.429V) = V \rho$

$0.429\rho_w= \rho$

The density of water under normal conditions is $1000kg / m ^ 3$ , so

$0.429(1000) = \rho$

$\rho = 429kg/m^3$

The density of the object is $429kg / m ^ 3$

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3 years ago

Question 6 (1 point)

jarptica [38.1K]

Explanation:

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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.

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3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

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3 years ago