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3 years ago

A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.

Physics

1 answer:

aleksandrvk [35]3 years ago

8 0

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

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What is the impulse of a 3kg object accelerating from rest to 12m/s?

ale4655 [162]

To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).

The answer is 36 kg m/s

6 0

3 years ago

During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=

Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

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6 0

3 years ago

In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 vo

andriy [413]

Answer:

r = 9.92 mm

Explanation:

Given that,

Mass of oil drop, $m=2\times 10^{-15}\ kg$

It acquires 2 surplus electrons, q = +2e $=3.2\times 10^{-19}\ C$

Potential difference, V = 620 V

Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

Since, E = V/r (r is distance between plates)

$mg=\dfrac{qV}{r}\\\\r=\dfrac{qV}{mg}\\\\r=\dfrac{3.2\times 10^{-19}\times 620}{2\times 10^{-15}\times 10}\\\\=0.00992\ m\\\\=9.92\ mm$

So, the distance between the plates is 9.92 mm.

6 0

3 years ago

If the gas inside the flask in the above exercise is cooled so that its pressure is reduced to a value of 715.7 torr, what will

evablogger [386]

Answer:

(a) 49.0 mm

Explanation:

#First we need to calculate the height of the mercury arm:

Reduced pressure is 715.7torr, Initial Pressure is 797.3torr, Required pressure is:

$P_g_a_s=P_a_t_m+P_h\\$

The difference in the two arms will give the pressure difference between the gas placed in the flask attached to an open-end mercury manometer.

$P_g_a_s=P_a_t_m+P_h\\797.3=P_a_t_m+(136.4mm-103.8mm)\\P_a_t_m=764.7mm\\\\h_t_o_t_a_l=136.4mm+103.8mm=240.2mm\\\\P_g_a_s=P_a_t_m+\bigtriangleup h\\\bigtriangleup h=P_g_a_s-P_a_t_m\\\\\bigtriangleup h=715.7mm-764.7mm\\=-49mm$

Hence the height of the mercury in the arm is 49.00mm

4 0

3 years ago

When the temperature of 0.0788 moles of a diatomic gas drops, its internal energy drops by -50.7 J. What was the temperature cha

miss Akunina [59]

Answer:

30.93

Explanation:

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5 0

3 years ago