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lys-0071 [83]
4 years ago
15

Which statement describes atomic radii as one moves from top to bottom within a group?

Physics
1 answer:
skad [1K]4 years ago
8 0

Answer:

a.The atomic radii increases from top to bottom within a group.

Explanation:

We know that

When we move from top to bottom in  a group then the atomic radii increases When we move from top to bottom in a group then number of shell increases.

Therefore, the distance of electron in outermost shell from the nucleus is also increase.Hence, the atomic radii  increases.

But when we move from left to right in a group then nuclear pull increases because number of electron increases in the same shell.

Therefore, size of atom decreases and atomic radii decreases.

Hence, the atomic radii increases as one moves from top to bottom within a group.

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A steam engine absorbs 4 x 105 J and expels 3.5 x 105 J in each cycle. What is its efficiency?
Wewaii [24]
Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J

From the first Law of thermodynamics, obtain useful work performed as
W = Qin  -  Qout
     = 0.5 x 10⁵ J

By definition, the efficiency is
η = W/Qin
   = 100*(0.5 x 10⁵/4 x 10⁵)
   = 12.5%

Answer: The efficiency is 12.5%
3 0
3 years ago
A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce
Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = MV_f-MV_i

where F is the force excerted by the wall, t is the time, M the mass of the ball, V_f the final velocity of the ball and V_i the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

 

3 0
3 years ago
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
4 years ago
Applications of pressure
Sunny_sXe [5.5K]
  • hydraulic press
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2 years ago
A ball is thrown upward with an initial velocity of +9.8 m/s. How high does it reach before it starts descending?
aksik [14]
Hope this helps you.

3 0
3 years ago
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