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3 years ago

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Who was the 43rd First Lady?

1 answer:

atroni [7]3 years ago

8 0

Answer: After serving as Second Lady from 1981 to 1989, Barbara Pierce Bush served as First Lady of the United States when her husband George H. W. Bush won the Presidency. She is also the mother of the 43rd President, George W. Bush, and of Florida's 43rd Governor, Jeb Bush.

Explanation:

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Give an example of a Claim.

Olin [163]

Answer:

Yeah

Explanation:

Look at the pic!!

5 0

3 years ago

What is the energy difference between parallel and antiparallel alignment of the z component of an electron's spin magnetic dipo

Readme [11.4K]

Answer:

$\Delta U= 4.8204\times10^{-24} J$

Explanation:

The difference between parallel and anti parallel alignment of the z component of an electron's spin magnetic dipole moment is given by

$U_1-U_2= (-\mu Bcos(\theta_2))-(-\mu Bcos(\theta_1))$

where μ= dipole moment B= strength of magnetic field and θ= angle between direction of magnetic field and dipole

here θ_2 and θ_1 are 180 and 0° respectively.

$U_1-U_2= (\mu B)-(-\mu B)$

$U_1-U_2= 2\mu B$

here μ is bohr magnetron or magnetic moment of an electron = 9.27×10^{-24}

put this value we get

$U_1-U_2= 2\times 9.27\times10^{-24}(0.26)$

$\Delta U= 4.8204\times10^{-24} J$

3 0

3 years ago

PLSSS HELP I WILL GIVE U BRAINLIEST annabelle and jose are building a complete circuit for their science inquiry in class they w

Agata [3.3K]

Answer:

i think its C hope this is rhight have a good day

Explanation:

6 0

3 years ago

Read 2 more answers

Nikolay [14]

Answer:

16

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

r' = 4r

So, the new force between the particles will be

$F'=\frac{kq_1 q_2}{(4r)^2}=\frac{kq_1 q_2}{16r^2}=\frac{1}{16}F$

So, the new force decreases by a factor of 16.

4 0

3 years ago

An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic f

meriva

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

$F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N$

Hence, this is the required solution.

3 0

3 years ago