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3 years ago

6

The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity Q

in a horizontal plane. Write down the equation of motion for a bead that is threaded on the rod, using the coordinates x and y of a frame that rotates with the rod (with x along the rod and y perpendicular to it). Solve for x (t). What is the role of the centrifugal force

1 answer:

Doss [256]3 years ago

7 0

Answer:

The answers can be found attached.

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What do you mean by population size explain

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Population size is the actual number of individuals in a population. Population density is a measurement of population size per unit area, i.e., population size divided by total land area. Abundance refers to the relative representation of a species in a particular ecosystem.

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3 years ago

How many possible combinations are there for the values of l and ml when n = 2?

Alexeev081 [22]

For n = 2, l will have two values 0,1

For l = 0, ml = 0

l = 1, ml = -1,0,1

These are the possible combinations.

The n is the principal quantum number

l is the azimuthal quantum number

ml is the magnetic quantum number.

The formula to find the azimuthal quantum number is

l = 0,1,2,3 .......n-1

If n = 2, the azimuthal quantum number,

l = 0 , 1

Then, the ml can be found using the formula,

ml = 2l +1

For l = 0 , ml = 2(0) + 1

= 1

Therefore, ml will have only one value, which is 0

For l =1 , ml = 2(1) + 1

= 3

Therefore, the ml will have three values, they are -1,0,1

Learn more about the quantum numbers in

brainly.com/question/16977590

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1 year ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

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4 years ago

The separation of positive and negative charges due to electric field is called a(n) ____.

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