All large radioactive atoms decay into smaller atoms by releasing alpha particles. Each alpha particle has 2 protons, 2 neutrons, and 0 electrons. The table below describes several neutral, low-mass atoms.

A 4-column table with 4 rows, labeled Stable Isotopes of Low-Mass Elements. The first column labeled element name has entries hydrogen, helium, lithium, beryllium. The second column labeled symbol has entries H, H e, L i, B e. The third column labeled atomic number has entries 1, 2, 3, 4. The fourth column labeled number of neutrons has entries 0, 2, 4, 5.

An alpha particle is also referred to as a nucleus of which isotope?

lithium-7
helium-4
hydrogen-2
helium-2​

Answers

Answer 1

Answer: Helium=4 is the answer

Explanation:

took the test and got it right that'

s the answer.

Answer 2

Answer:

It's b

Explanation:

I took the quiz on EDGE and got it right.


Related Questions

A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.

Answers

Answer:

The answer  is The acceleration is double its original value.

Explanation:

It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.

Hope this helps....

Have a nice day!!!!

Answer:

The acceleration is half of its original value

Explanation:

A small cylinder is rolled along a ruler and completes two revolutions. The circumference is the distance around the outside of a circle. What is the circumference of the cylinder? A 4.4 cm B 5.2 cm C 8.8 cm D 10.2 cm

Answers

Answer; 4.4cm

Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm

distance traveled =10.2-1.4=8.8

since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so

2x2πxr =8.8

so the circumference will be 2πr=4.4cm

 

Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. When astronomers make such a statement, how can they know what was happening inside galaxies way back then

Answers

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

Read more at; https://brainly.com/question/15995216

What is the value of work done on an object when a 0.1x102–newton force moves it 30 meters and the angle between the force and the displacement is 25°?

Answers

Answer: A. 2.7 x 10^2 joules

Explanation: I’m sorry for the guy above me!

in how many ways can five basketball players be placed in three postitions?

Answers

Answer:

Well if they playing a game like that

Will mark as BRAINLIEST.......

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation 4x³+3x²-5x+2 , where x is in meters and t is in sec.
a)Find velocity of particle at i) t=2 sec ii) t=4 sec.
b) Find the acceleration of the particle at t=3 sec.

Answers

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

[tex]x=4t^3+3t^2-5t+2[/tex]

Where,

x is in meters and t is in sec

We know that,

Velocity,

[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]

(a) i. t = 2 s

[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]

At t = 4 s

[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]

(b) Acceleration,

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]

Pu t = 3 s in above equation

So,

[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²

an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil​

Answers

Answer:

1) The density of the object = 2500 kg/m³

2) The density of the oil = 1250 kg/m³

Explanation:

1) The information relating to the question are;

The mass of the object in air = 0.250 kgf

The mass of the object in water = 0.150 kgf

The mass of the object in the oil  = 0.125 kgf

By Archimedes's principle, we have;

The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced

The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf

∴ The weight of the water displaced = 0.1 kgf

Given that the object is completely immersed in the water, we have;

The volume of the water displaced = The volume of the object

The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)

The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³

The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³

The density of the object = 2500 kg/m³

2) Whereby the mass of the object in the oil = 0.125 kgf

The upthrust of the oil = The weight of the oil displaced

The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil

The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf

The weight of the oil displaced = The upthrust of the oil

Given that the volume of the oil displaced = The volume of the oil, we have;

The volume of the oil displaced = 0.0001 m³

The mass of the 0.0001 m³ = 0.125 kg

Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.

The density of the oil = 1250 kg/m³.

Forensic toxicologist analyze and identify drugs that are confiscated from criminals



True
False

Answers

The answer should be false. Because the drugs are not confiscated from criminals

A missile is moving 1350 m/s at a 25° angle it needs to hit a target 23,500 m away in a 55° direction in 10.2 seconds what is the magnitude of its final velocity

Answers

Answer:

3504 m/s

Explanation:

Let x be the horizontal component of distance

y - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity

horizontally: x = V_x ×t + ½×a_x×t²  

plugging the values we get

23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²  

⇒ax = 19.2 m/s²  

Moreover,

V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s  

similarly in vertical direction:

y = V_y×t + ½×a_y×t²  

23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²  

⇒a_y = 258 m/s²  

Also,

V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s  

Therefore

V = √(V'x² + V'y²) = 3504 m/s  

therefore,  magnitude of final velocity of missile=3504 m/s

THANKS  

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Will mark as BRAINLIEST..... A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated at a height of 100m. How long does it take the packet to reach the ground ? What is it's final velocity ?

Answers

Answer:

PFA

:-)

Explanation:

A box with mass of 2 kg is pushed directly horizontally over a horizontal surface (with friction) at a constant speed of 10 m/s. The force of the push is 60 N. How much thermal energy is generated pushing the box a distance of 15 m

Answers

Answer:

E= 600 W

Explanation:

Given that

m = 2 kg

Speed ,  v= 10 m/s

Force , F= 60 N

Given that box is moving with constant velocity, it means that friction force will be 60 N.

f = 60 N

Therefore total energy generated

E= f x v

E= 60 x 10 = 600 W

E= 600 W

Thus the answer will be 600 W.

(b) A cylinder of cross-sectional area 0.65m2 and
height 0.32m has a mass of 2. Ikg. If there is a
cavity inside, find the volume of the cavity.
(Density of cylinder = 11.053 kg/m^3)​

Answers

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]

Step Four:

calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]

Recall that the cavity has the same height as the original cylinder

[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]

Can you solve this question please help me with this​

Answers

Answer:

Explanation:

The velocity ratio of  a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;

VR = R/r

Since radius = diameter/2

VR = (D/2)/(d/2)

VR = D/d

D is the diameter of the wheel and 'd' is the diameter of the axle.

Given VR = 3 and d = 5cm

3 = D/5

D = 15 cm

If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.

b) Workdone by the load = Load * distance moved by load

Given load = 60kg

Distance moved by load = 2π*radius of axle

Distance moved by load  = 2π(0.025) = 0.157

workdone by load = 60* 0.157 = 9.42J

Effort = Workdone by load/distance moved by the wheel

Effort = 9.42/2π(0.075)

Effort = 9.42/0.471

Effort = 20kg

Hence the effort applied is 20kg

c) MA = Load/Effort

MA = 60/20

MA = 3

d) Efficiency = MA/VR * 100%

Efficiency = 3/3 * 100%

Efficiency = 100%

Atoms of the same element will always have the same number of Question Blank but will have different numbers of Question Blank if their mass numbers are different.

Answers

Answer:

proton and neutron respectively.

Explanation:

Atoms of the same element will always have the same number of proton but will have different numbers of neutron if their mass numbers are different.

The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive

Answers

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

What did Bohr's model of the atom include that Rutherford's model did not have?
a nucleus
energy levels
electron clouds
smaller particles

Answers

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

Answer:

energy levels

hope this helped!

Explanation:

A student is planning an investigation on the properties of different types of matter. What would be the best method to find the volume of an irregularly shaped object, such as a rock?

Answers

Explanation:

Volume is the amount of space an object takes up, while mass is the amount of matter in an object. ... To find the volume of an irregular sized object, one would use the displacement method for measuring volume and place the object in water and measure the amount of water that is displaced.

Answer:

To measure the volume of an irregularly shaped object, pour some water in a measuring cylinder. Then suspend the irregularly shaped object with a thread. After that , move the object gradually downwards and immerse it in water. The volume of the irregularly shaped object is the difference between the volume of the liquid before and after. After measuring the difference, we come to know about the volume of the irregularly shaped object.

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 23.5 m/s at an angle of 39.5∘ above the horizontal. When the softball leaves her hand, it is 11.5 m above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.

Answers

Answer:

66.86m

Explanation:

Velocity of ball thrown, u = 23.5 m/s

Initial height of the ball above the water, H = 11.5 m

Angle of projection, θ = 39.5°

Vertical components of veloclty = usinθ

Horizontal components of veloclty = ucosθ

The soft ball hits the water after time 't'

Considering the second equation of motion

S = ut + 1/2at^2........ 1

But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:

H = ut +/- 1/2gt^2

H = - 11.5m

U = usinθ

θ = 39.5°

a = -g = -9.8m/s^2

- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2

-11.5m = 23.5(0.6360)t - 4.9t^2

-11.5m = 14.946t - 4.9t^2

4.9t^2 -14.946t-11.5m = 0

Since the ball drifted horizontally

D = (Ucosθ)t

Where θ = 39.5°

U = 23.5m/s t=

Alternatively,

horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s

now how long does it take the ball to raise to a peak and fall to the water.

vertical component of velocity = 23.5 sin 39.5º = 14.947m/s

time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec

peak reached above cliff top is

h = ½gt² = ½(9.8)(1.5252)²

= ½×22.797

= 11.3985m

now the ball has to fall 11.3985+ 11.5 = 22.8985m

time to fall from that height is

t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec

add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869

now haw far does the ball travel horizontally in that time

d = vt = 18.1331 ×3.6869= 66.856m

= 66.86m

Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?

Answers

this process is called parellelogram method of resolving vectors.

Need help finding the average speed.

Answers

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.381.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/30.95 + 1.61 + 0.56 = 3.12 / 3 = 1.040.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?

Answers

Answer:

Rounded to three significant figures:

(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:

A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].Maxima

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].

The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].

Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].

Minima

The dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.

First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].

For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].

Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233 MeV. The masses of the proton and the deuteron are 1.007 276 467 u and 2.013 553 212 u, respectively. Based on this data, what is the mass of the neutron

Answers

Answer:

Explanation:

Energy of gamma ray = 2.2233 MeV

Let mass of neutron be n  amu  

mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .

in terms of energy this mass defect will be equal to energy of gamma ray

1 amu = 931 MeV

931 [ 2.013553212 - ( 1.007 276 467 + n ) ] =  2.2233

( 1.007 276 467 + n ) -  2.013553212  = .00238807733

n = 1.008664822 amu

so mass of neutron = 1.008664822 amu

The acceleration due to gravity near Earth ... Select one: a. varies inversely with the distance from the center of Earth. by. varies inversely with the square of the distance from the center of Earth. c. is a constant that is independent of altitude d. varies directly with the distance from the center of Earth.

Answers

Answer:

b. varies inversely with the square of the distance from the center of Earth.

Explanation:

Comparing the Newton's law of universal gravitation and second law of motion;

from Newton's second law of motion,

F = ma ............. 1

from New ton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex] ........... 2

Equating 1 and 2, we have;

mg =  [tex]\frac{GMm}{r^{2} }[/tex]

g = [tex]\frac{GM}{r^{2} }[/tex]

Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.

The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points

Answers

Answer:

c

Explanation:

The marginal cost curve image  has been attached from which we can clearly, indicate that

ATC = average total cost

AFC = average fixed cost

AVC = average variable cost.

From the graph we can indicate that the marginal cost curve

(c) Lies above the AVC curve when the AVC curve slopes downward.

Is there a way for us to control motion

Answers

Answer:

They are:

1) change position

2) distract yourself

3) Get fresh air

4) Face the direction you are going.

5) Drink water.

6) Play music.

7) Put your eyes on horizon.  

Explanation:

Hope it helps.

Which value would complete the last cell?

(1 point)

3.0

100.0

25.0

4.0

Answers

Answer:

4.0

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as

Force (F) = mass (m) x acceleration (a)

F = ma

With the above formula, we can obtain th acceleration of the body as follow:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

F = ma

20 = 5 x a

Divide both side by 5

a = 20/5

a = 4 m/s²

Therefore, the value that will complete the last cell in the question above is 4.

At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counselors measure the two "dry" legs of a right triangle. What is the length in meters of the swimming course in the figure below?

Answers

Answer:

47.17 m

Explanation:

From the diagram of the question attached, The length of the legs are 25 m and 40 m . This legs form a right angle triangle with the length of the swimming course (L).

Pythagoras theorem states that for a right angle triangle with hypotenuse a and legs b and c, then:

a² = b² + c²

In the triangle, the length of the swimming course (L) is the hypotenuse and the two legs are 25 m and 40 m. Using Pythagoras:

L² = 25² + 40²

L² = 625 + 1600

L² = 2225

L = √2225

L = 47.17 m

A cat stalks an unsuspecting mouse. The movement of the cat is shown in the following graph of the horizontal position, x, against time, t. What is the average speed of the cat between t = 2s and t = 6s?

Answers

Answer:

0.25 m/s

Explanation:

Average velocity = change in displacement / change in time

v = (1.5 m − 2.5 m) / (6 s − 2 s)

v = -0.25 m/s

The average speed is 0.25 m/s.

Answer:

0.25 m/s

Explanation:

khan academy

SHOW ADEQUATE WORKINGS IN THIS SECTION

12. Wale and Lekan of 55kg and 60kg respectively ran a race of 200m.

i. Calculate their work done in KJ

ii. If wale finished the race in 25secs while Lekan finished in 30secs, calculate their

power and who is more powerful out of these two. ( g = 10m/secs)

13. A machine has a velocity ratio of 5 and is 800

/0 efficient. If the machine is carrying a load

of 200kg, what will be effort applied?


Please help me answer anyone that you understand ​

Answers

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force was applied?

Answers

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

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