different between group IA and IIA elements
Answer:
Group IA elements have only one valency electron while Group IIA have two valency electrons.
Group IA elements have cations with higher charge density hence polarizing anions easier resulting into covalent character while Group IIA elements have cations with lower charge density hence difficulty in distorting anions resulting into a ionic character. This is due to difference in cationic radii and charges
14. A 3.0 kg metal ball, at rest, is hit by a 1.0 kg metal ball moving at 4.0 m/s. The 3.0 kg ball moves
forward at 2.0 m/s and the 1.0 kg ball bounces back at 2.0 m/s.
(a) What is the total kinetic energy before the collision?
Answer:
The kinetic energy before the collision was 8 J
Explanation:
The given parameters are;
The mass of the metal ball at rest, m₁ = 3.0 kg
The velocity of the ball at rest = 0 m/s
The mass of the metal ball that hits the one at rest, m₂ = 1.0 kg
The velocity of the metal ball that hits the one at rest, v₂ = 4.0 m/s
The velocity with which the 3.0 kg ball moves forward = 2.0 m/s
The velocity with which the 1.0 kg ball bounces back = 2.0 m/s
a) Kinetic energy, K.E. = 1/2 × m × v²
Where;
m = The mass of the of the object in motion
v = The velocity of the object
Before the collision, we have;
Initial K.E. = 1/2 × m₁ × v₁² + 1/2 × m₂ × v₂²
∴ Initial K.E. = 1/2 × 3.0 kg × (0 m/s)² + 1/2 × 1.0 kg × (4.0 m/s) ² = 8 J
The kinetic energy before the collision = The initial K.E. = 8 J.
Do you attract the earth or the earth attract you ?Which one is attracting with a large force ?You or the earth
i need help finding the calculation for the bed forces if diagrams.
Explain how the aperture geometry relates to the
diffraction pattern.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
how the aperture geometry relates to the
diffraction pattern:
Diffraction is the spreading out of waves as they pass through an aperture or around objects.it occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave.For every small aperture sizes,the vast majority of the wave is blocked.For large apertures the wave passes by or through the obstacle without any significant of diffraction.
in an aperture with width smaller than the wavelength,the wave transmitted through the aperture spreads all the way around the behave like a point sources of waves.
single slit diffraction pattern
The diffraction pattern made by waves passing through a slit of width [tex]\alpha[/tex] (larger than∫) can be understood by imagining a series of point sources all in phase along the width of the slit.The waves moving directly forward are all in phase,so they from a large central maximum.
if the waves travels at an angle Ф from the normal to the slit,then there is a path difference x between the waves production at the two end of the slit.
x=a sinΦ
The path difference between the top and middle waves is λ/2 then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.)at the angle so there is no resultant wave at this angle.Thus a minimum is the diffraction pattern is obtained at
λ=α sinθ
Now slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two section ,which is identical to the result of pairing off matching sources in the bottom two sections.in this case we obtained from the minimum.
λ/2=α/4 sinθ
we can divided the slit aperture into six equal sections and pair off sources in the top two divisions and then the bottom two,to give destructive interference for every matched pair.The minimum of intensity are obtained at angles
nλ = α sinθ
where n is an integer (1,2,......), but not n=0.There is a maximum of intensity in the center of the pattern. This process only gives the position of the minima,does not work for positions of the maxima,and so does not give the intensities of the maxima.
Jack cannot run as fast as Paul. He decides to go at his own pace and not run with Paul. Which principle of fitness is he following?
Progression
Overload
Individuality
Reversibility
Answer:
individuality
Explanation:
since he cannot run with paul he decided to run by himself making his exercise individual
Answer: individuality
Explanation:
(you can ignore this its just the deffintion of individuality)
This is a crucial principle, the fundamental fact that everyone is different! Everyone responds to training in a different way. If you are walking or cycling with a friend, and doing exactly the same amount of training, don’t be concerned if one of you gets fitter faster than the other – this is what individualisation is all about.
It might be that one of you is having some pressure at work or difficulties at home, but wherever it is, it’s surprising what can affect your training. Some days your training can go really well and the next day, even though it was exactly the same length workout, it can be a nightmare. This is individualisation.
You get a job delivering water. You calculate how much work is done picking up each 20 L bottle of
water and raising it vertically 1 m. For every 100 bottles you deliver, you will use Select.... (g =
9.8 m/s2)
-196J
-2,000 J
-19,600 J
-196,000J
Answer:
The work done by picking up 100 20-L bottles and raising it vertically 1 meter is 19614 joules.
Explanation:
By the Work-Energy Theorem, the work needed to raise vertically 100 bottles of water is equal to the gravitational potential energy, units for work and energy are in joules:
[tex]\Delta W = \Delta U_{g}[/tex] (1)
Where:
[tex]\Delta W[/tex] - Work.
[tex]\Delta U_{g}[/tex] - Gravitational potential energy.
The work is equal to the following formula:
[tex]\Delta W = n\cdot \rho \cdot V \cdot g \cdot \Delta h[/tex] (2)
Where:
[tex]n[/tex] - Number of bottles, dimensionless.
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta h[/tex] - Vertical displacement, measured in meters.
If we know that [tex]n = 100[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]V = 0.02\,m^{3}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta h = 1\,m[/tex], then the work done is:
[tex]\Delta W = (100)\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot (0.02\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1\,m)[/tex]
[tex]\Delta W = 19614\,J[/tex]
The work done by picking up 100 20-L bottles and raising it vertically 1 meter is 19614 joules.
can someone please explain some calculations on vernier calliper
Answer:
Hey mate here's your answer ⤵️
Vernier caliper least counts formula is calculated by dividing the smallest reading of the main scale with the total number of divisions of the vernier scale.LC of vernier caliper is the difference between one smallest reading of the main scale and one smallest reading of vernier scale which is 0.1 mm 0r 0.01 cm
Hope it was helpfulllStanding waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest
Answer:
The rank of the frequencies from largest to smallest is
The largest frequency of oscillation is given by the string in option D
The second largest frequency of oscillation is given by the string in option B
The third largest frequency of oscillation is given by the string in option A
The smallest frequency of oscillation is given by the string in option C
Explanation:
The given parameters are;
The mass per unit length of all string, m/L = Constant
The tension of all the string, T = Constant
The frequency of oscillation, f, of a string is given as follows;
[tex]f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}[/tex]
Where;
T = The tension in the string
m = The mass of the string
L = The length of the string
n = The number of overtones
[tex]Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K[/tex]
For the string in option A, the length, L = 27 cm, n = 3 we have;
[tex]f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K[/tex]
For the string in option B, the length, L = 30 cm, n = 4 we have;
[tex]f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K[/tex]
For the string in option C, the length, L = 30 cm, n = 3 we have;
[tex]f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K[/tex]
For the string in option D, the length, L = 24 cm, n = 4 we have;
[tex]f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K[/tex]
Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;
1 ) [tex]f_D[/tex] 2) [tex]f_B[/tex] 3) [tex]f_A[/tex] 4) [tex]f_C[/tex]
The order of the frequencies is [tex]f_D>f_B>f_A>f_C[/tex]
Standing waves:The frequency of the standing wave in a string tied at both ends is given by:
[tex]f=\frac{nv}{2L}[/tex]
where n is the mode of frequency
v is the velocity of the wave
and L is the length of the string.
Now the velocity of a wave in a string tied at both ends is given by
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where T is the tension and μ is the mass per unit length.
Since T and μ are the same for all the strings, velocity [tex]v[/tex] will be the same for all.
Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.
[tex]f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}[/tex]
Hence, [tex]f_D>f_B>f_A>f_C[/tex]
Learn more about standing waves:
https://brainly.com/question/1698005?referrer=searchResults
Please help me out
please show solvings
Answer:
5) The approximate height is about 80 m. Choice (c)
6) The stone takes about 3 seconds to rise to its maximum height
Explanation:
Free Fall Motion
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the gravitational acceleration, which value is [tex]g = 9.8 m/s^2[/tex].
The distance traveled by a dropped object is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
Question 5
Given the stone reaches the ground in t=4 seconds, the height of the tower is:
[tex]\displaystyle y=\frac{9.8*4^2}{2}=78.4\ m[/tex]
The approximate height is about 80 m. Choice (c)
Question 6
Vertical Motion
The vertical motion of an object is controlled by the force of gravity. This means that there is a non-zero net force acting on the object that makes it accelerate downwards.
If the object is thrown upwards at speed vo, its speed at time t is:
[tex]v_f=v_o-g.t[/tex]
The stone reaches its maximum height when the final speed is zero, thus:
[tex]v_o-g.t=0[/tex]
Solving for t:
[tex]\displaystyle t=\frac{v_o}{g}[/tex]
The stone is thrown vertically upwards with vo=30 m/s, thus:
[tex]\displaystyle t=\frac{30}{9.8}[/tex]
[tex]t=3.06\ s[/tex]
[tex]t\approx 3\ s[/tex]
The stone takes about 3 seconds to rise to its maximum height
When going from a fast speed to a slow speed how is light bent?
1.away from the normal line
2.toward the normal line
Answer:
Toward the normal line
Explanation:
I had the same question on a quiz
Answer with explanation!!!
Answer:
The light enters the box along the normal to the side of the box or perpendicularly to the box's surface
Therefore, the light is expected to travel straight through the box
However, when a rectangular glass prism is placed inside the box, the light can then be refracted to pass through the box in the given path as shown
Light bends towards the normal when passing from a less dense medium to a denser medium with larger refractive index and vice versa
Therefore, when the glass prism with a larger refractive index than air is inclined with the top part further away from the incident beam and the bottom part closer to the incident beam as shown, the refracted ray through the box will be shifted downwards as shown in the drawing created with Microsoft Visio
Explanation:
[tex]Refractive \ index, n = \dfrac{sin(i)}{sin(r)}[/tex]
Where;
i = The angle between the incidence light and the normal line
r = The angle between the refracted light and the normal line
n = The refractive index
When n > 1 is large, we have, ∠i > ∠r and the light is bent towards the normal when moving from a less dense medium to a denser medium and vice versa.
Plzz Help easy quesion for 20 points Why does someone get burned if they touch an old filament bulb that has been on for a few hours? Plzz Help
Answer:
In reality, the filament gets so hot it in a real sense bubbles off molecules and electrons. Now and again this material gathers as a dull spot at the highest point of the bulb. Eventually, the filament falls apart, gets frail, and breaks, subsequently finishing the life of the light. Lights radiate light by siphoning an electric flow through a dainty tungsten fiber. The filament warms and emits light. Over the long haul, the filament oxidizes and turns out to be increasingly fragile, until it splits up and the bulb goes out. ... Tungsten picks up obstruction as it warms.
Hope this helped :)
The dancing bear family loves when their trainer gives them little treats to reward them for a good performance. If the trainer gives the dancing bear family 34 treats each show, how many treats will the trainer need for 22 shows?
Answer:
748 treats.
Explanation:
If the trainer gives out 34 treats and gave them out for 22 shows, then to find the total you need to multiply 34 by 22, or (a longer but more simple way) add 34, 22 times.
My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and stay another 60 minutes. We can ______________________ to see about how long we have to remain at the surface to have enough no stop time. (choose all that apply)
Answer:
1) Periodically check the no stop or NDL time on their computers
2) The dive computer planning mode can be used if available
3) Make use of a dive planning app
4) Check data from the RDP table or an eRDPML
Explanation:
The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop
The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning
The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.
A 45kg sled is being pulled from camp by 5 dogs each capable of exerting 25N force on the sled. If the sled starts from rest and the frozen ground exerts 15N of friction, how far will the sled be from the camp after 7s?
Answer:
58.8m
Explanation:
Define one joule heat.
Answer:
The joule is the standard unit of energy in electronics and general scientific applications. One joule is defined as the amount of energy exerted when a force of one newton is applied over a displacement of one meter. One joule is the equivalent of one watt of power radiated or dissipated for one second.
Hope it helps ^^
PLEASE ANSWER ASAP! It would be EXTREMELY appreciated if you could answer both. These are free responses, so try to make your answer somewhat original. Thanks in advance! And I will mark brainliest!
Why does a third class lever cannot magnify force?
Explanation:
The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.
A swing held up by someone at one side has the potential to swing when released.
True
False
2. __________is the pressure exerted by the layer of air around the Earth
Answer:
Atmospheric Pressure
Explanation:
Assume that Michael's teacher has a rule that if a student talks out-of-turn three times in one day, that student must stay in for recess for the
rest of the week. Which of the following methods is she using to control student behavior?
a. folkway
b. sociobiology
c. informal sanction
d. formal sanction
Imagine you've recently insulated your loft. Give a reason why your heating bill may still increase.
Answer:
follow me for the answer
AP physics here! Please help me answer only this question and then explain how you got your answer. After that, I will do the rest of the questions by myself using your explanation. Come here only if you know AP physics.
Answer:
displacement: 0
Distance: 30
Explanation:
lol, i asked my brother, he's also in 10th grade
A girl runs at a speed of 3.9 m/s off a high dive and hit the water 1.8 s later.
a. How high was the diving board?
b. How far horizontally was she from the board when she hit the water?
c. If she had just dropped off the board, would her time to drop to the water been longer, shorter or the same?
Answer:
(a) the height of the diving board is 22.896 m
(b) the horizontal distance traveled by the girl is 7.02 m
(c) if she had just drop off the board, her time to drop to the water would have been longer.
Explanation:
Given;
initial speed of the girl, u = 3.9 m/s
time to hit the water, t = 1.8 s
(a) the height of the diving board is calculated as;
h = ut + ¹/₂gt²
h = (3.9 x 1.8) + ¹/₂ x 9.8 x 1.8²
h = 7.02 + 15.876
h = 22.896 m
(b) the horizontal distance traveled by the girl is calculated as;
X = ut
X = 3.9 x 1.8
X = 7.02 m
(c) if she just drop off the board, then the initial speed will be zero;
h = ut + ¹/₂gt²
h = 0 + ¹/₂gt²
2h = gt²
[tex]t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2 \ \times\ 22.896 }{9.8} }\\\\t = 2.16 \ s[/tex]
Thus, if she had just dropped off the board, her time to drop to the water would have been longer.
A car accident my rolls off a cliff as it leaves the cliff it has a horizontal velocity of 13 ms it hiys the ground 60 m from the shoreline calculate the height of the cliff
Answer:
104.59 m
Explanation:
The following data were obtained from the question:
Horizontal velocity (u) = 13 ms¯¹
Horizontal distance (s) = 60 m
Height (h) =?
Next, we shall determine the time taken for the car to get to the ground. This can be obtained as follow:
Horizontal velocity (u) = 13 ms¯¹
Horizontal distance (s) = 60 m
Time (t) =?
s = ut
60 = 13 × t
Divide both side by 13
t = 60 / 13
t = 4.62 s
Finally, we shall determine the height cliff. This can be obtained as follow:
Time (t) = 4.62 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
h = ½gt²
h = ½ × 9.8 × 4.62²
h = 4.9 × 21.3444
h = 104.59 m
Thus, the height of the cliff is 104.59 m
1. How much heat is produced if you heat .25 kg water from 25 degrees C to 100 degrees C. The specific heat of water is 4180 J/kg C.
Answer:
Q = 78375 [J]
Explanation:
To solve this problem we must use the following ideal equation for the thermal energy of an element or substance depending on the temperature change.
[tex]Q=m*C_{p}*(T_{final}-T_{initial})[/tex]
where:
Q = heat [J]
m = mass = 0.25 [kg]
Cp = specific heat = 4180 [J/kg*C]
Tinitial = 25[°C]
Tfinal = 100 [°C]
[tex]Q =0.25*4180*(100-25)\\Q = 78375 [J][/tex]
A 747 requires a takeoff speed of 270 meters per second if it is to get off the
ground. Calculate the acceleration required for the 747 that begins from rest
to takeoff on a 1000 meter long runway.
Answer:
a = 36.45[m/s²]
Explanation:
We can calculate the acceleration value by means of the following expression of kinematics.
[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]
where:
vf = final velocity = 270 [m/s]
Vo = initial velocity = 0 (begins from rest)
a = acceleration [m/s]²
x = distance required by the plane = 1000 [m]
[tex](270)^{2} =0+2*a*1000\\72900=2000*a\\a=36.45 [m/s^{2} ][/tex]
I need help in question 9 plzzz
True is the correct answer
marked as brainiest if correct
Answer:
I think the the answer is creating and layout and template style ( C )
Explanation:
I did it before and i was checking my notes and i wrote that down , Hope this Helps :)