A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuured in a scale?

1. ) It's good to remember contact information such as phone number or email.

2.) Your address, so if they need to mail anything to you, they can. This is also a necessary point of information on most job applications anyway.

3.) Past job experiences and ways to contact them. This is probably one of the most important things as it shows you've worked before and are willing to talk to past employers.

May I have brainliest please? :)

Answer:

the other answer is correct

Explanation:

i do flvs and this was on the career final exam

Answer:

I'm gonna say it's D

Explanation:

but when u do the experiment on in u head you'll actually find out that it is actually , indeterminable

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

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.

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.

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.

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.

Answer:

Explanation:

The density of gold is 19.3 g/cm³. Thus, you can determine the density of the unknown cube by doing the following. Get a measuring cylinder (marked in cm³) that contains a certain volume of water (preferably above average but not close to been filled up). Get a weighing balance (that can read in grams) also.

Measure the mass of the unknown stone (using the weighing balance) and record. Take the initial volume of the water in the measuring cylinder and record (in cm³) and then drop the unknown stone inside the measuring cylinder gently (avoid splashes). Record the final volume of the cylinder after the unknown stone was dropped.

Then calculate the density of the stone by using the formula; mass ÷ change in volume

The change in volume can be determined by; Final volume - initial volume

If the answer obtained from the calculation (of the density of the stone) is not around 19.3 g/cm³ (say 19.3 ± 0.2), then the stone is not gold but if it is around 19.3 g/cm³, then the stone is gold.

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

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

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 ^^

A boy of mass 85 kg is riding a bicycle in a circle with a radius of 7.2 meters, and it takes 6 seconds to make one revolution, then his velocity will be equal to 7.54 m/s.

The movement of an object along the circle's edge which path is always at the tangent to any specific point on the circle is described by tangential velocity.

As a result, tangential velocity is the amount of motion along a circle's edge that may be measured at any given time.

The speed at any location that really is tangent to the rotating wheel is quantified as the tangential velocity in a circular motion. The connection between rotational acceleration and tangential velocity is expressed by the formula. The part of movement along the circle's edge that can be detected at any time is called tangential velocity.

The given data in the question,

Radius, r = 7.2 meters.

Time, t = 6 seconds.

v = 2πr / t

v = 2π(7.2)/6

v = 7.54 m/s.

Therefore, the velocity of the boy is 7.54 m/s.

To know more about tangential velocity:

https://brainly.com/question/28738284

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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]

Answer:

displacement: 0

Distance: 30

Explanation:

lol, i asked my brother, he's also in 10th grade

Answer:

In biology, cell theory is the historic scientific theory, now universally accepted, that living organisms are made up of cells, that they are the basic structural/organizational unit of all organisms, and that all cells come from pre-existing cells. Cells are the basic unit of structure in all organisms and also the basic unit of reproduction.

Explanation:

The three tenets to the cell theory are as described below:

All living organisms are composed of one or more cells.

The cell is the basic unit of structure and organization in organisms.

Cells arise from pre-existing cells.

There is no universally accepted definition of life. Some biologists consider non-cellular entities such as viruses living organisms,[1] and thus reasonably disagree with the first tenet. Throughout this article, it will lead you through the history of cell theory, how the discovery of cells was made possible, what the cell theory has become today and background information and history regarding other opposing concepts of cell theory.

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 :)

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]

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

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.

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.

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.

Answer:

is it 3?

Explanation:

Im taking a guess and just dividing 6 and 2

Answer:

Explain. yes it is possible because when the net force is zero and acceleration is zero. ... no, Neglecting air friction, the only force acting on this object is the force of gravity downward. There is no force of equal magnitude as the force of gravity acting in the opposite direction.

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.

Answer:

58.8m

Explanation:

Answer:

Atmospheric Pressure

Explanation:

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

True is the correct answer

Answer:

Two moles of aluminum chloride [tex](AlCl_3)[/tex] are produced when six miles of hydrogen Chloride [tex](HCl)[/tex] react with plenty of aluminum

Explanation:

6 Moles of [tex]HCl[/tex] will only react with 2 moles of [tex]Al[/tex] irrespective of the number of moles of each compound present. The reaction wiil take place in this ratio only. The products produced will be 2 moles of [tex]AlCl_3[/tex] and 3 moles of [tex]H_2[/tex] this ratio will also be constant.

So, six moles of hydrogen chloride [tex](HCl)[/tex] will react with plenty of aluminum to produce many 2 moles of aluminum chloride [tex](AlCl_3)[/tex].

Answer:

dark-field microscopy

Explanation:

A darkfield microscope can be regarded as a brightfield microscope with a significant modification to condenser. There is an opaque disk, with a diameter of about 1cm placed in between the illuminator and condenser lens, the disk responsible for the blocking of most of the light from the illuminator, passing through the condenser to objective lens , which will then form hollow cone of light that has its focus on the specimen. Using a darkfield microscope the light that get to the objective is the only refracted or reflected light of structures in the specimen. And the image formed usually display a bright object with a dark background.

It should be noted that darkfield microscope is a type of microscopy works by allowing only light waves that have reflected from or refracted though the sample to enter the lens system.