Chadwick received the Doctor of? 1.Human Letters

What is a structure that organizes motion of chromosomes

strojnjashka [21]

<span>Cytoplasm: <span>the entire contents of the cell, exclusive of the nucleus and bounded by the plasma membrane.

Hope this helped. :)</span></span>

4 0

3 years ago

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A uniform, thin, solid door has height 2.30 m, width 0.875 m, and mass 24.0 kg. (a) find its moment of inertia for rotation on i

vekshin1

A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary? So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it. So basically I have no idea where to even start with this.

so A

3 0

3 years ago

A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude r

iren [92.7K]

The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

learn more about oscillators brainly.com/question/15169199

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8 0

1 year ago

A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r

Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1: Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0

3 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

3 years ago