Give the relationship between the number of valence electrons in an atom's
valence electron shell and the position of the element on the Periodic Table
Answer:
they're reactions
Explanation:
The relationship between the valence electrons and position is: the number of valence electrons determines the position
What is valence electron?This is the number of electrons in the outermost shell of an atom.
NOTE: The outermost shell is called valence shell
Position in Periodic tableThis is where an element is located in the periodic table
Relationship between valence electrons and positionThe position of an element in the periodic table is determined by the number of valence electrons.
For example
Sodium, Na (atomic number of 11) has the following electronic configuration
1st shell = 2 electrons2nd shell = 8 electrons 3rd (valence) shell = 1 electronSince the valence electron is 1, thus, sodium is located in group 1 of the periodic table.
Thus, we can see that the position of an element in the periodic table is related to the valence electron(s) in the atomic shell of the element.
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Before we make measurements, let's make sure we understand the circuit. 1. Select all of the following that correctly describe what a volt meter and ammeter measure. Select all that apply: A volt meter measures the potential difference (or voltage) across a circuit element. A volt meter measures the potential difference (or voltage) passing through a circuit element. A ammeter measures the electric current passing through a circuit element. A ammeter measures the electric current across a circuit element.
Answer:
the correct answers are a and c
Explanation:
In an electrical circuit there are two important quantities to measure, such as voltage and current.
Voltage is the potential difference between two points in a circuit
current is the number of electrons you pass through a given point per unit of time.
Now let's analyze each answer
a) true. The potential difference across an element
b) False. The potential difference is u field there is no physical entity that moves
c) True. The current is electrons in motion and these pass through the given element
d) False. There is a physical quantity that passes through the point
the correct answers are a and c
1.A body of mass 10kg falling freely was found to be falling at a rat of 20m/s what force will stop the body in 2second?
Answer:
50N
Explanation:
force it is falling with can be found by mass into acceleration and then devide by half to find force that could stop it in 2 sec
Question 1 of 10 Which characteristic of nuclear fission makes it hazardous?
O A. The process destroys every atomic nucleus that goes through it.
O B. The process changes existing atoms into atoms of different elements
C. The process requires very high pressures.
O D. The radioactive wastes it produces can spread to any part of the environment
Answer:
D.
The radioactive wastes it produces can spread to any part of the environment.
Explanation:
Fission is the name given to the splitting of a heavy nucleus into two medium-mass nuclei. Fission is said to be induced if it results from the absorption of a neutron to form an unstable nucleus that decays by fission. The induced fission usually produces further neutrons accompanying the fission fragments.
Answer: D. The radioactive wastes it produces can spread to any part of the environment
Explanation:
Why do astronomers use frequencies other than the visible ones when they are
investigating the universe?
The magnetic field due to a utility wire is 0.10 mT when you are at a distance of 10 meters from it. What current (in Amperes) flows through the wire?
Answer:
I = 5000 A
Explanation:
We will use Ampere's Law to calculate the current:
[tex]B = \frac{\mu I}{2\pi r}\\\\[/tex]
where,
B = Magnetic Field Strength = 0.1 mT = 1 x 10⁻⁴ T
μ = Permeability of Free Space = 4π x 10⁻⁷ N/A²
I = Current = ?
r = radius = 10 m
Therefore,
[tex]1\ x\ 10^{-4}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(I)}{2\pi(10\ m)}\\\\I = \frac{(1\ x\ 10^{-4}\ T)(2\pi (10\ m))}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]
I = 5000 A
A wave has a frequency of 30Hz and wave length of 40cm. What is the velocity of the wave?
Answer:
12m/s
Explanation:
v=fλ
30×(40÷100)=
12m/s
1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane
Answer:
the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J
Explanation:
Given that the data in the question;
angle of inclination with respect to the ground [tex]\theta[/tex] = 30°
length of plane d = 2m
m₁ = 1 kg
m₂ = 0.5 kg
now, velocity of the first block at midpoint;
[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
v² = gsin[tex]\theta[/tex]d
v = √( gsin[tex]\theta[/tex]d)
g is 9.8 m/s
so we substitute
v = √( 9.8 × sin30° × 2)
v = √( 19.6 )
v = 3.13 m/s
Now, velocity just after collision of the blocks will be;
(m₁ + m₂)v₂ = m₁v
v₂ = m₁v / (m₁ + m₂)
we substitute
v₂ = (1 × 3.13) / (1 + 0.5)
v₂ = 3.13 / 1.5
v₂ = 2.0866 m/s
now, final kinetic energy will be;
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²
we substitute
[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]
[tex]KE_f[/tex] = 7.35 + 3.2654
[tex]KE_f[/tex] = 10.62 J
Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J
A woman has one damaged fallopian tube. The damage completely blocks the opening of the tube at the ovary. How will this most likely affect her fertility?
Her likelihood of conceiving will be reduced by at least 50%.
Her eggs, if fertilized, will not implant properly.
Her eggs will not mature inside the ovary.
Her chances of conceiving twins will be doubled.
Answer:Her likelihood of conceiving will be reduced by at least 50%.
Explanation:
EDGE 2020
Answer:
Her likelihood of conceiving will be reduced by at least 50%
Explanation:
took the test on edge, got it right
: A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks
Answer:
r = 0.0173 m = 1.73 cm
Explanation:
Here, the centripetal force of the block will be providing the required breaking tension in the string:
[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]
where,
r = radius = ?
m = mass of block = 0.13 kg
v = tangential spee of block = 4 m/s
T = Breaking Strength = 30 N
Therefore,
[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]
r = 0.0173 m = 1.73 cm
When finding the radius of the string at the point it breaks, the tangential
velocity is assumed to be constant.
The radius when the string breaks is [tex]\underline{6.9 . \overline 3 \times 10^{-3}} \ m[/tex]Reasons:
The mass of the small block, m = 0.130 kg
Initial radius of the circle of rotation = 0.800 m
Tangential velocity, v = 4.00 m/s
The radius of the path of rotation is reduced as the string is pulled
Breaking strength of the string = 30.0 N
Required:
The radius of the circle when the string brakes
Solution:
[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]
Where;
r = The radius of the circle of rotation
When the string brakes, w have;
Centripetal force = Breaking strength of the string = 30.0 N
Which gives;
[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]
The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]
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A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding. a) up the incline, b) down the incline ?
Answer:
Hope It Help
Explanation:
That's all I know
Please Help!!!
For A&B
Answer:
A. 40N
B. 5m/s
Explanation:
A.
Impulse is equal to the area under the curve of a force vs. time graph. In this case, the area is in the shape of a triangle with base 8 (12-4=8) and perpendicular height 10:
Area of a triangle = (1/2)bh
A=(1/2)*8*10
=40
ANSWER: 40N
B.
Impulse = mass * velocity
40 = 8v
v = 5
ANSWER: 5m/s