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QveST [7]
3 years ago
14

The outermost shell is referred to as the __________ shell. the electrons in the outermost shell are also referred to as the ___

_________ electrons.
Physics
1 answer:
Llana [10]3 years ago
6 0
The outermost shell is known as the valence shell. The electrons in this level are known as valence electrons.
You might be interested in
. 30
schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0
2 years ago
Which word in the sentence is a helping verb Jamie could probably repair his bike by himself
Vesna [10]
Helping verbs come before the main verb, the main verb in the sentence is repair therefor the helping verb would be could.
6 0
2 years ago
Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
Debora [2.8K]

493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

5 0
3 years ago
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
2 years ago
1.) A stone falls from rest from the top of a cliff.
KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

     0 = 21.56t - 71.148

t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

or

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0
2 years ago
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