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3 years ago

Jim Abbott was drafted ________ overall in the first round of the 1989 draft.

Physics

2 answers:

Rom4ik [11]3 years ago

7 0

The answer is "eighth". And he was drafted in 1988 not 1989.

cupoosta [38]3 years ago

4 0

He was drafted 8th overall

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How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?

Luda [366]

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375 x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

4 0

3 years ago

Which statement is TRUE? A) In a series circuit, the resistors do not share the current. B) In a series circuit, as light bulbs

3241004551 [841]

The answer is B, actually did the research

7 0

3 years ago

Read 2 more answers

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

3 years ago

A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0

3 years ago

Help meee please fill in the blanks !!

Ratling [72]

The atomic number is the simply the number of protons in the atom. So in the first row with atomic number 2, the number of protons is 2.

If the atom has no charge, which I think you can assume for all of these, the number of electrons is equal to the number of protons. So the number of electrons is also 2.

The number of neutrons (which are the particles with no charge in the nucleus) is simply the mass number minus the atomic number i.e. 4 - 2 = 2.

The isotopic symbol is the symbol which is found on the periodic table of elements. There are 2 numbers associated which each element on the table. The larger is the mass number and the smaller is the atomic number. The atomic number or number of protons is what identifies the element. Looking at the periodic table ( https://sciencenotes.org/wp-content/uploads/2015/01/PeriodicTableOfTheElementsBW.pdf or https://simple.wikipedia.org/wiki/Periodic_table_(big) ), it can be seen that the element on the first row above with an atomic number of 2 is Helium with a symbol He. The number that is included with the name is simply the mass number which is 4 in this case, which tells us that this type of helium has 2 neutrons.

Another type (or isotope) of helium is Helium-3 which has one neutron.

Try the next row and post back if you have trouble with it

3 0

3 years ago