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3 years ago

Northern lights & auroras can be seen within the what sub layer

1 answer:

4 0

Auroras and northern lights can be seen within the ionosphere. This is the range where cosmic and the sun's radiation are ionized.

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3 PHYSICAL SCIENCE: In order to figure out what is in a sealed box, you perform some tests. The box seems heavy for its size and

rjkz [21]

Some sort of magnetic metal

Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials

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3 years ago

In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.

babymother [125]

The answer is B) <span>equilibrium
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5 0

3 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m

Nezavi [6.7K]

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current $I = 2.5 \times 10^{4}$ A

Time $t = 40 \times 10^{-6}$ sec

We know that the current is the rate of flow of charge.

From the formula of current,

Where $Q =$ charge transfer between cloud and earth.

$Q =I t$

$Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}$

$Q = 1$ C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

3 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$