Answer:
0.6192
Explanation:
1.44 x 0.430=0.6192
the temprerture of an object is 40°c.what is this tempreture in farhenhite scale?
This is an exercise of thermal scales.
Data:T = 40 °C
T = °F ¿?
To convert degrees Celsius to degrees Fahrenheit, we have the formula:
[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]
We substitute our data into the formula.
[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]
First we divide 9/5 multiplied by 40.
[tex]\bf{=75+32}[/tex]
Add 75 plus 32
[tex]\bf{=104 \ ^{\circ}F}[/tex]
Therefore, the temperature change of 40 °C on the °F scale is equal to 104.
[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]
an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of the object?
The height of the object will be -5.19 cm
A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.
Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall
So let,
v = Image distance from the mirror = -33.5 cm
u = object distance from the mirror (concave) = 24 cm
hi = Image height = 7.25 cm
h = height of the object = ?
Using below formula to find height of the object
-v/u = hi/h
Putting all value in the formula we get
-(-33.5)/(-24) = 7.25/h
h = -5.19 cm
Therefore the height of the object will be -5.19 cm
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A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m. Calculate the impulse given to the bàll by the floor
The impulse given to the ball by the floor is 0.2865 kg.m/s.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.
The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s
The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s
Substitute the values into the expression, we get
Impulse = m(v- u)
Impulse=0.5 x (4.852- 5.425 )
Impulse = - 0.2865 kg.m/s
Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.
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An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air resistance is negligible,calculate the initial velocity of the arrow and the height of the building?
Need an answer urgently please
The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is;
u is the initial velocity of fall = ? m/sec
h is the distance of fall = 200 m
g is the acceleration of free fall = 9.81 m/sec²
H is the height of the building
t is the time period = 10 second
According to Newton's second equation of motion,
[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]
- ve shows the direction is downward.The magnitude of the initial velocity is found as;
u = 29.05 m/sec
The height of the building
[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]
Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
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A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m
bCalculate the speed of the ball just before a it hits the gound.
Answer:
Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].
Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)
Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].
Explanation:
Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.
Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.
Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.
Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.
Rearrange this equation to obtain an expression for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].
For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612
(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) angular frequency = velocity / radius
ωf = 10/2 = 5 rad/s and ωi = 20 rad/s
Angular acceleration, α = ωf - ωi /t
Put the values, we get
α = -15/5 = -3 rad/s²
Coefficient of friction, μ = a/g = rα/g
Plug the values, we get
μ = 0.612
Thus, the coefficient of friction corresponding to the frictional force is 0.612.
(c) The energy lost = heat generated
energy lost = 1/2 Iω² + 1/2 Mv²
energy lost = 1/2 MR²ω² + 1/2 Mv²
Plug the values, we get
energy lost = 7500 J
Thus, the increase in the thermal energy is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands
s= (v- 2gH) sin2θ /g
s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81
s = 7.78m
Thus, the horizontal distance is 7.78 m
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Water's specific heat capacity is 4.184 J/(g · °C). How does this compare to the heat capacities of most common substances?
Answer:
Explanation:
Comment
It's high. It takes quite a bit of heat to get water to change its heat content. Most substances don't reach much more the 0.5 J/(gram*oC)
Question 3 of 15
Which statement describes an advantage of using nuclear fission to produce
electricity?
A. It uses a fuel made of atoms with large nuclei that will be
destroyed.
B. It is much more efficient than any other process used to generate
power.
C. It happens in large buildings that are expensive to build.
D. It produces radioactive waste products that must be stored safely.
Answer:
it is much more efficient than any other process to generate power
Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air moving with a wind speed of 73.0 mi/h in a hurricane? Assume the air is 51.2 km from the center of the hurricane "eye."
The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.
Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h
We have to find the magnitude of the angular momentum
Let,
ρ = Density of air = 1.29 kg/m^3
v = Speed of wind = 73.0 mi/h = 0.032 km/s
M = angular momentum of air
Let the volume of air be 1 m^3
Mass = Volume x ρ = 1 x 1.29 = 1.29 kg
Momentum = M = mass x velocity
Momentum = 1.29 x 0.0032
Momentum = 4.128 x 10^(-3) kg·m^2/s
Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
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What is the mass of an object if it has 915J of energy when it is 6.50m above the ground?
A.851kg
B0.0696kg
C.583kg
D.14.4kg
Answer:
D
Explanation:
GPE = mgh
GPE / gh = m
m = GPE/gh
m = 915 / 6.5 x 9.8
m = 915 / 63.7
m = 14.4kg
A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike. (b) how long will take to do that?
Answer:
a) 0.2m/s
b) 5m/s
Explanation:
a) acceleration=∆v/∆t
v=10
t=50
10/50=1/5
=0.2m/s
b) time= d/s
d=50m
s=10
50/10
=5m/s
Answer:
See below
Explanation:
Average speed = (0+ 10)/2 = 5 m/s
then to cover 50 m will take 50 m / 5 m/s = 10 seconds
change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2
A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed when it strikes the ground below? (Use conservation of energy.)
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
Time of motion of the projectileThe time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
h is height of the cliffv is velocityt is time of motion265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
Final velocity of the projectilevyf = vyi + gt
where;
vyf is the final vertical velocityvyi is initial vertical velocityvyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
vxf is the final horizontal velocityvxi is the initial horizontal velocityvxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
vf is the speed of the projectile when it strikes the ground belowvf = √(149.322² + 130.8²)
vf = 198.51 m/s
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Part A
Calculate the spring constant.
• Look at the graph of x vs. t graph. (You may want to double-click on
it to examine it in the Data Tool view.)
. From the graph, determine the period T.
• The first video frame lists the mass.
• Use the equation below to solve for the spring constant k. Show
your work below.
m
T = 2π√√T
Answer:
Using 3 periods to get an accurate reading:
3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s
m = 0.250 kg
Using algebra:
k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec
What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the ground?
-441J
735J
-735J
441J
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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1. If the amplitude of a vibrating mass attached to the free end of an ideal spring is
25 cm, then the mass in one period travels 'a distance:
a) 1m
b) 0
c) 25m
The mass in one period travels 'a distance less than 25 cm which will be 0.
What is amplitude of a pendulum?
The amplitude of a pendulum is the maximum displacement of the pendulum.
If the maximum displacement of the pendulum is 25 cm, the mass in one period travels a distance less that 25 cm.
Thus, then the mass in one period travels 'a distance less than 25 cm which will be 0.
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Which of the following can be contracted from contact with bloodborne pathogens?
HIV can be contracted from contact with bloodborne pathogens.
Other bloodborne diseases are HBV, malaria, syphilis and brucellosis
What are bloodborne pathogens?Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.
Blood borne pathogens can also be contacted through the following means
Se- xual contactNeedle contactIn conclusion; HIV can be contracted from contact with bloodborne pathogens.
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loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room. Explain how the Iron Heating element S loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room . Explain how the Iron Heating element S
Answer:
suna bro i am nepali and nepali are pro
The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.
The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.
What is gravitation?Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.
By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.
m = 250 N / 9.8 m/s²
m = 25.51 kg
Multiply the acquired mass by Pluto's gravitational acceleration (g);
W = (25.51 kg) x (0.61 m/s²)
W = 15.56 N
Thus, the item will only be 15.56 N in weight.
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Answer: B.
15.6 newtons
Explanation: edmentum
A 3,204 kg tree positioned on the edge of a cliff 247 m above the ground breaks away and falls into the valley below which is considered zero potential energy. If the tree’s mechanical energy is conserved, what is the speed of the tree just before it hits the ground in meters/sec?
Let's see
PE is turned to KE as per law of conservation of energy[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]
[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]
[tex]\\ \rm\Rrightarrow v²=4940[/tex]
[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]
NGC 300 is the name of a spiral galaxy that looks similar to our Milky Way galaxy. It contains billions of stars. How is NGC 300 different from our solar system?
NGC 300 is different from our solar system because of its smaller size.
How is NGC 300 different from our solar system?NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.
So we can conclude that NGC 300 is different from our solar system because of its smaller size.
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A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on it. State with a reason what is observed about the motion of the trolley.
Answer:
Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.
A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)
The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
u is the initial velocity of canonball= 47.4 m/sec
g is the acceleration of free fall = 9.81 m/sec²
v is the velocity after 1.23 s
According to Newton's first equation of motion,
[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]
Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
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In a recent study of how mice negotiate turns, the mice ran around a circular 90∘ turn on a track with a radius of 0.10 m. The maximum speed measured for a mouse (mass = 18.5 g) running around this turn was 1.39 m/s. Part A What is the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed?
The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
Minimum coefficient of frictionThe minimum coefficient of friction is calculated as follows;
F = μmg = mac
μmg = mac
where;
ac is centripetal accelerationμg = ac
μg = v²/r
μ = v²/rg
where;
μ is the minimum coefficient of frictionv is velocityr is radius of the pathg is acceleration due to gravityμ = (1.39²)/(0.1 x 9.8)
μ = 1.97
Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
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I was having trouble converting this p-t graph into a velocity-time graph and acceleration-time graph. Can someone convert it and send a picture of it?
We can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.
What is a velocity time graph?The velocity time graph shows the change of velocity with time. Recall that velocity is the change of position with time.
As such, when we have a position time graph as shown, we can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.
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Aude slipped on ice going 7.6 m/s .What was her velocity before she hit the wal?
Answer:
Define universal gravitational constant. Why is newton's law of gravitation called universal law?
Resolve the vector shown below into its components.
Answer:
D. s=3x+4y
Explanation:
The line is at the 3rd column in the 4th row.
n Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.
What is the concentration of H+ ions at a pH = 2?
mol/L
What is the concentration of OH– ions at a pH = 2?
mol/L
What is the ratio of H+ ions to OH– ions at a pH = 2?
: 1
At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².
What is pH?The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.
The concentration of H⁺ ions is calculated as:
pH = -log [H⁺]
2 = -log [H⁺]
[H⁺] = 10⁻²
The concentration of OH⁻ ions is calculated as:
pH + pOH = 14
pOH = 14 - 2
pOH = 12
Solving further:
pOH = -log [OH⁻]
12 =-log [OH⁻]
[OH⁻] = 10⁻¹²
At a pH = 2, the ratio of hydrogen ions to hydroxide ions:
10⁻² ÷ 10⁻¹² = 10¹⁰
Therefore, at pH = 2 the ratio is 10¹⁰.
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Using Hooke's law, what happens when the displacement is tripled?
If the displacement is tripled then the spring force also is tripled.
What is Hooke's law?The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).
Froom the hooks law;
F = kx
Where,
F is the applied force
x is the displacement in case 1
Case 2;
x'=3x
F'=kx'
F'=3KX
F'=3x'
Where,
x' is the displacement when the force is tripled
F' is the force in the case2
Hence, if the displacement is tripled then the spring force also be tripled.
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DON'T ADD LINKS!
Imagine that a single cell below goes through four cell divisions.
How many cells are produced? To find out, draw the parent cell
and cells present after each division. Then write the number of
cells in the space below each drawing. (Hint: the number of cells
doubles with each division.)
Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.
How many cells are produced?When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.
So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.
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