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2 years ago

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notes of diffusion

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ivolga24 [154]2 years ago

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4<br> Explica las características principales de las fuer-<br> zas de acción y reacción.

Naya [18.7K]

-Surgen de una interacción.
-Nunca aparece una sola: son dos y simultáneas.
-Actúan sobre cuerpos diferentes: una en cada cuerpo.
-Nunca forman un par de fuerzas: tienen la misma línea de acción.
-Un cuerpo que experimenta una única interacción no está en equilibrio, pues sobre el aparece una fuerza unica que lo acelera. Para estar en equilibrio se requieren por lo menos dos interacciones.

Las mas importantes son la 2,3,4 característica

5 0

3 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

What is the mass of a cannonball if the force a force of 2500 N gives the cannonball an acceleration of 200 m/s^2??

vampirchik [111]

The answer is a.12.5kg because i just did the test and it was correct.

hope this helps

5 0

3 years ago

6. If two objects experience the same net force, but they have different masses, which object will accelerate at

Digiron [165]

Answer:

i would say the heavier object

4 0

3 years ago

Read 2 more answers

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

3 years ago