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2 years ago

Two objects 5 kg and 7 kg are attached to the end of inextensible string which passes over frictionless pulley

Physics

1 answer:

Bumek [7]2 years ago

3 0

Answer:

ans:

tenson(T) = 20 N

acceleration (a) = 2.86 m/s

Explanation:

T + mg = Mg

T = Mg - mg

T = g( M - m )

T = 10× ( 7-5 )

T = 20 N

again;

T = 20

Ma = 20

a = 20 / 7

= 2.86 m/s

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La masa de un objeto en la tierra es de 100 kg.

andreev551 [17]

Answer:

a) 980 N

b) 100 kg

c) 163.33 N

Explanation:

The weight of the object on Earth is the product of its mass times the acceleration due to gravity, that is:

$W=m\,g=(100\,\,kg)\,g=(100\,\,kg)\,(9.8\,\,m/s^2)=980\,\,N$

The mass of the object on the moon, is exactly the same (100 kg) since mass is an inherited value of the object.

Considering that the acceleration of gravity on the Moon is about 1/6 that we have on earth (g), then the weight of the object on the Moon becomes:

$W=m\,\frac{g}{6} =\frac{980}{6}\,\,N=163.33\,\,N$

3 0

3 years ago

What does the "coefficient of friction" tell you?

Arte-miy333 [17]

a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

8 0

3 years ago

Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

barxatty [35]

For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles

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: