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4 years ago

In the thomson model of the atom, where was the positive charge located?

1 answer:

7 0

I don't know if you still need this, but here's an answer anyways.

It was a giant cloud of matter, with negatively charged electrons inside.

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Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

Explanation:

Gravity would be exerting on the 2 masses

$G_1 = Mg = 15.3*9.81 = 150.093N$

$G_2 = mg = 8.29*9.81= 81.32N$

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

$F = G_1 - G_2 = 68.77N$

This force would generate torque on the solid pulley

$T = FR = 68.77 * 0.0075 = 0.5158 Nm$

We can also calculate the pulley moments of inertia, with it being solid

$I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2$

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

$\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2$

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

$\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad$

(c)The time it takes to get to this distance is

$\theta = \frac{\alpha t^2}{2}$

$t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513$

$t = \sqrt{0.513} = 0.716s$

(b)The final angular speed of the disk is

$\omega = \alpha t = 1300.58*0.716 = 931 rad/s$

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

$v = \omega R = 931*0.0075 \approx 7m/s$

5 0

3 years ago

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

Bond [772]

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω $L_{total}$ where ω is angular frequency

= 2πf $L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0

3 years ago

A black hole can be considered a star that has...

joja [24]

B is the answer...

mark brainliest

3 0

3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )