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4 years ago

Where on the table would a group’s ionization energy be the greatest?

Physics

1 answer:

ANEK [815]4 years ago

6 0

Explanation:

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

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A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the

Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

$\Delta{K} + \Delta{U} = 0$

$\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0$

Simplifying the above equation, we get

$v^2 = 2G\dfrac{M}{r}$

We can rewrite this as

$v^2 = 2\left(G\dfrac{M}{r^2}\right)r$

Note that the expression inside the parenthesis is simply the acceleration due to gravity $g$ so we can write

$v^2 = 2gr$

where $v$ is the launch velocity.

6 0

3 years ago

A 19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a ho

Anton [14]

Answer:

d) 700 m/s

Explanation:

if k is the force constant and x is the maximum compression distance, then:

the potential energy the spring can acquire is given by:

U = 1/2×k×(x^2)

and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy = final energy

Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

Ki = Uf

1/2×m×(v^2) = 1/2×k×(x^2)

m×(v^2) = k×(x^2)

v^2 = k×(x^2)/m

= (500)×((21×10^-2)^2)/(19×10^-3 + 8)

= 2.75

v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

m1×v1 = v×(m1 + M)

v1 = v×(m + M)/(m1)

= (1.66)×(19×10^-3 + 8)/(19×10^-3)

= 699.86 m/s

≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.

5 0

3 years ago

PLEASE HELP!!! When you lift a book from the ground to your desk, what kind of work do you do, negative or positive? By lifting

Snowcat [4.5K]

When someone lifts a book from the ground, the work you use is positive. By lifting the book, you change it's energy and it's original place The book gains, kinectic energy.

Hope I helped.

8 0

3 years ago

Linear expansivity?<br>

shusha [124]

Linear expansivity, area expansivity and volume or cubic expansivity are

7 0

3 years ago

Consider two stars that are identical (same size, temperature and luminosity). Star A is 10 pc away from us and Star B is 20 pc

Mkey [24]

Answer:

Star A would be brighter than Star B

Explanation:

The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.

A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.

Brightness and distance from earth have an inverse square relationship.

$brightness$ ∝ $\frac{1}{distance^{2} }$

Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.

If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be $\frac{1}{2^{2} }$ of Star A.

Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.

3 0

4 years ago