Answer:
Sports drink will show increase in specific gravity
While
Water will decrease in SG
Explanation:
This is because the relative density of urine will increase because of presence of ions in the sport drink unlike in water that has very little ions
Leticia timed how fast five apple slices turned brown (oxidate) after being being dipped in different preservatives such as lemon juice, fruit freshener, salt water and lime soda. another part of the experiment had the apple slices simply set out without any chemical on them. all parts of the experiment had the apple slices in the same indoor conditions such as humidity temperature and lighting; also only one variety of apple-red delicious was used.
Identify each of the following independent variable, dependent variable, constants, control experiment and repeated trials.
Answer:
Independent Variable: Choice of preservative
Dependent Variable: Time it took to brown
Controlled Variables/Constants: Climate, Size of apple slices, and amount of preservative on each slice
Control: Apple Slices without preservatives on them
Repeated Trials: One?
Explanation:
What is the opposite of 4/4
Answer: -1 or -4/4
Explanation:
4/4 simplifies to 1 and the opposite of 1 is -1.
Calculate the following expression: 2.36 + 3.38 + 0.355 + 1.06 =
the answer wil be 2.36+3.38+0.355+1.06=7.155
A parallel plate capacitor consists of two square parallel plates separated by a distance d. the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be % of the energy stored
B) There will be % of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
Answer:
E) the energy stored will quadrupled
Explanation:
The correct question is
A parallel plate capacitor consists of two square parallel plates separated by a distance d. If i double the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be 1/4 of the energy stored
B) There will be 1/2 of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
The initial energy stored in the capacitor [tex]E_{i}[/tex] = [tex]\frac{1}{2}CV^{2}[/tex]
where C is the capacitance
V is the potential difference
If I double this voltage, while holding every other parameters constant, the new energy stored will be
[tex]E_{n}[/tex] = [tex]\frac{1}{2}C(2V)^{2}[/tex] = [tex]\frac{4}{2}CV^{2}[/tex]
[tex]E_{n}[/tex] = [tex]2CV^{2}[/tex]
dividing new energy stored by the initial energy stored, we have
[tex]E_{n}/E_{i}[/tex] = [tex]2CV^{2}[/tex] ÷ [tex]\frac{1}{2}CV^{2}[/tex] = 4
the energy stored will be quadrupled.
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
1.25cm
Explanation:
Using
Minimum, as dsinစ = (m+1/2) lambda
Third dark fringe m= 2
dsinစ = (2+1/2)lambda
d(y/L)= (5/2) lambda
Y= 5/2* lambda *L/d
So substituting
=[ (500E-9m)(5m)/0.5E-3] 5/2
=0.0125m
= 1.25cm
Explanation:
Suppose you attempt to pour out 100 g of salt, using a pan balance for measurements, while an elevator that is accelerating upward. Will the quantity (weight) of salt be too much, too little, or the correct amount? Explain
Answer:
the correct amount
Explanation:
The pan balance has two pans . When the elevator is accelerating upward , the apparent weight of both objects placed on either pan increase . The net effect is that they cancel out the effect of accelerating elevator . Hence the apparent weight of salt remains same as and equal to its real weight.
In case of spring balance, its apparent weight would have increased . Its apparent weight would have measured more than its true weight.
masses of 3kg on a smooth horizontal table.It is connected by a light string passing at the edge of the table to another mass of 2kg hanging vertically.When to the system is released from rest with what acceleration do the mass move
Answer:
aaawwwwwwwsssaaaasasss
A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444 s. Determine the distance from the surface of the earth to the surface of the moon. Note: The speed of light is 2.9979 ? 108 m/s.
Answer:
7.92 × 10^8 M
A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444s. then the distance from the surface of the earth to the surface of the moon would be 3.9632×10⁸ meters.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
As given in the problem A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444s
the given speed of the light is 2.9979 ×10⁸ m/s
As we know that
Distance =speed ×time
Distance = 2.9979 ×10⁸× 2.6444
=7.92644× 10⁸
This is the total distance covered by both ways from the moon to earth
The actual distance of the moon from earth would be half of this distance
Thus, the distance from the surface of the earth to the surface of the moon would be 3.9632×10⁸ meters.
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The critical period is a special time of development in humans. True False
Answer:
True
Explanation:
Without proper development during the critical period, the growth of some life skills may be severely delayed or stunted.
During which stage of the water cycle does water from the ocean form clouds?
Answer:
Condensation
Explanation:
Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
Answer:
This process is called condensation.
Explanation:
When a cloud becomes full of liquid water, it falls from the sky as rain or snow.
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston
Answer:
The force is [tex]F_1 = 400.8 \ N[/tex]
Explanation:
From the question we are told that
The first diameter is [tex]d_1 = 4.0 \ cm = 0.04 \ m[/tex]
The second diameter is [tex]d_2 = 8.0 \ cm = 0.08 \ m[/tex]
Generally the first area is
[tex]A_1 = \pi * \frac{d^2_1 }{4}[/tex]
=> [tex]A_1 = 3.142 * \frac{0.04^2}{4}[/tex]
=> [tex]A_1 = 0.00126 \ m^2[/tex]
The second area is
[tex]A_2 = \pi * \frac{d^2_2 }{4}[/tex]
[tex]A_2 = 3.142 * \frac{0.08^2}{4}[/tex]
[tex]A_2 = 0.00503 \ m^2[/tex]
For a hydraulic press the pressure at both end must be equal .
Generally pressure is mathematically represented as
[tex]P = \frac{F}{A}[/tex]
=>
[tex]\frac{F_1}{A_1 } = \frac{F_2}{A_2 }[/tex]
=> [tex]F_1 = \frac{1600}{0.00503} * 0.00126[/tex]
=> [tex]F_1 = 400.8 \ N[/tex]
Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells. If a seagull drops a shell from rest at a height of 16 m, how fast is the shell moving when it hits the rocks
Answer:
v = 17.71 m / s
Explanation:
We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity
v² = v₀² - 2 g (y -y₀)
v² = 0 - 2g (y -y₀)
when it hits the stone the height is zero and part of the height of the seagull I
v² = 2g y₀
v = Ra (2g i)
let's calculate
v =√ (2 9.8 16)
v = 17.71 m / s
Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from source B has an intensity of 20 μW/m2. This is all we know about the two beams. Which of the following statements about these beams are correct? A) Beam B carries twice as many photons per second as beam A. B) A photon in beam B has twice the energy of a photon in beam A. C) The frequency of the light in beam B is twice as great as the frequency of the light in beam A. D) A photon in beam B has the same energy as a photon in beam A. E) None of the above statements are true.
Answer:
A) Beam B carries twice as many photons per second as beam A.
Explanation:
If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.
We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.
With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.
If the period of a simple pendulum is T and you increase its length so that it is 4 times longer, what will the new period be?
Answer:
T' = 2T
Explanation:
The time period of a simple pendulum is given by the relation as follows :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of the pendulum
g is acceleration due to gravity
If the length is increased four time, new length is l' = 4l
So,
New time period is :
[tex]T'=2\pi \sqrt{\dfrac{l'}{g}}\\\\T'=2\pi \sqrt{\dfrac{4l}{g}}\\\\T'=2\times 2\pi \sqrt{\dfrac{l}{g}}\\\\T'=2\times T[/tex]
So, the new time period is 2 times of the initial time period.
A 24.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?
Answer:
18.83m/s
Explanation:
In the first 11meters, we can calculate the Kinectic energy since we know that
the floor is frictionless then work done by the horizontal force= Kinectic energy gain by the block in the first 11m.
Then Kinetic Energy = wordone
But work done= Force × distance
Kinetic Energy=(225×11)= 2475J
In the next 10.0 m the coefficient of friction is 0.20, the Kinectic energy has equal value to the difference in workdone by both the horizontal and frictional force
K.E= (Force× distance) - ( mass × gravity× coefficient of friction × distance at 10 m)
K.E= [(225×10)-(.20× 9.8×10× 24)]
K.E= 2250-470.4
= 1779.6J
The addition of the Kinectic energy above give us the total Kinectic energy experience by the crate.
Total Kinectic energy= 1779.6+2475
= 4254.6J
the final speed of the crate after being pulled these 21.0 m?
Total distance= 11m + 10m = 21 m
Then the final speed can be calculated from the total Kinectic energy, since we know that
K.E= 0.5mv^2
V= √(2K.E/m)
= √(2×4254.6)/24
Final speed v = √354.55
Final speed v= 18.83m/s
Therefore, the final speed of the crate after being pulled these 21.0 m is 18.83m/s
Show that (a)KE=1/2mv2
Answer:
[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}[/tex]
Explanation:
[tex] \underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}[/tex]
Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :
[tex] \sf{W = FD}[/tex]
⇒[tex] \sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)[/tex]
⇒[tex] \sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}[/tex]
⇒[tex] \sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }[/tex]
⇒[tex] \sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}[/tex]
The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²
∴ [tex] \sf{KE= \frac{1}{2} m {v}^{2} }[/tex]
[tex] \sf{ \underline{ \bold{ {proved}}}}[/tex]
Hope I helped!
Best regards!!
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3-). If the cell is powered by a 12.0-V battery and has a resistance of 1.30 , how long does it take for a 0.133-mm layer of silver to build up on the teapot? (The density of silver is 10.5 multiplied by 103 kg/m3.)
13.4. Young's modulus for iron is 1.9 x 10' Pa. When an iron wire 1.0 m long with a cross-sectional area of 4.0 mm
supports a 100 kg load, the wire stretches by
(
Answer:
0.0129 m
Explanation:
ΔL = FL / (EA)
where ΔL is the deflection,
F is the force,
L is the initial length,
E is Young's modulus,
and A is the cross sectional area.
F = mg = 100 kg × 9.8 m/s² = 9800 N
A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²
ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))
ΔL = 0.0129 m
A Pipe Filled with Helium A certain organ pipe, open at both ends, produces a fundamental frequency of 293 Hz in air Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 293 Hz in air. How does your answer to Part A, fHe, change
Complete question:
A certain organ pipe, open at both ends, produces a fundamental frequency of 293 Hz in air. If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.97 g/mol and the molar mass of helium to be 4.00 g/mol. Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 293 Hz in air. How does your answer to first question, fHe, change?
Answer:
The fundamental frequency that helium produced is 788.6 Hz.
Since the fundamental frequency of air in closed pipe is still the same value as open pipe, the fundamental frequency of helium will not change.
Explanation:
The speed of sound in pipe when filled with air is given by;
[tex]v = \sqrt{\frac{\gamma RT}{M} }[/tex]
Where;
γ = 1.4
R = 8.31 J/mol.K
M = 0.02897 kg/mol
T = 20°C = 293 K
[tex]v_{air} = \sqrt{\frac{1.4* 8.31*293}{0.02897} } = 343 \ m/s[/tex]
The speed of sound in pipe when filled with helium is given by
[tex]v_{He} = \sqrt{\frac{1.4* 8.31*293}{0.004} } =923.14 \ m/s[/tex]
Now, determine the fundamental frequency Helium will it produce
v = fλ
[tex]\frac{V_{air}}{F_o{air}} = \frac{V_{He}}{F_o{He}}\\\\F_o{He} = \frac{F_o{air}*V_{He}}{V_{air}} \\\\F_o{He} = \frac{(293)*(923.14)}{343}\\\\F_o{He} = 788.6 \ Hz[/tex]
Since the fundamental frequency of air in closed pipe is still the same value as open pipe, the fundamental frequency of helium will not change.
PLEASE SEND GIVE THE ANSWER AS FAST AS POSSIBLE
Q1) State similarities & difference between the laboratory thermometers & the clinical thermometers
Q2) Give two examples for conductors & insulators.
Q3) Give reason: Wearing more layers of clothing during winter keeps
us warm
Answer:
(1): Similarities Both thermometers are used to measure temperature and both of them use mercury,Differences Clinical thermometer is used to measure human body temperature whereas laboratory thermometer is used to measure temperature of other object which has higher temperature than human body temperature
(2)Examples of conductors include metals, aqueous solutions of salts, Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
(3)Air acts as insulator of heat. This layer prevents our body heat to escape in the surroundings. More layers of thin clothes will allow more air to get trapped and as a result we will not feel cold. So wearing more layers of clothing during winter keeps us warmer than wearing just one thick piece of clothing
Explanation:
A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in air is 340 m/s. How many beats per second do the people on the fire engine hear
Answer:
The values is [tex]f_b =14.9 \ beats/s[/tex]
Explanation:
From the question we are told that
The speed of the fire engine is [tex]v = 5\ m/s[/tex]
The frequency of the tone is [tex]f = 500 \ Hz[/tex]
The speed of sound in air is [tex]v_s = 340 \ m/s[/tex]
The beat frequency is mathematically represented as
[tex]f_b = f_a - f[/tex]
Where [tex]f_a[/tex] is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as
[tex]f_a = [\frac{v_s + v }{v_s -v} ]* f[/tex]
substituting values
[tex]f_a = [\frac{340 + 5 }{340 -5} ]* 500[/tex]
[tex]f_a = 514.9 \ Hz[/tex]
Thus
[tex]f_b =514.9 - 500[/tex]
[tex]f_b =14.9 \ beats/s[/tex]
What is the mass of erath
Average speed is calculated by dividing distance traveled by
time. How is average velocity calculated?
Answer:
The average velocity of an object is its total displacement divided by the total time taken. In other words, it is the rate at which an object changes its position from one place to another. Average velocity is a vector quantity. The SI unit is meters per second.
hope this helps you
Answer:
displacement divided by time
Explanation:
A small, dense ball is launched from ground level at an angle of 50° above the horizontal. The ballâs initial speed is 22 m/s, and it lands on a hard, level surface at the same height from which it was launched. The ball then bounces and reaches a height of 75% its peak height it achieved at launch. Assume air resistance is negligible.(a) Find the maximum height reached by the ball during its first parabolic arc. (b) Find the distance between the launch point to where the ball lands the first time. (c) Find the distance between the launch point to where the ball lands the second time?
Answer:
a) Hmax = 10.86 m
b) Rx = 48.64 m
c) Rx¹ = 36.48m
Explanation:
Given that Ф = 50°
v₀ = 22 m/s
a)
hmax = v₀²sin²Ф / 2g
hmax = 22² × sin²50 / 2 × 9.8
hmax = 14.49 m
Hmax = 75% of hmax
Hmax = 0.75 × 14.49
Hmax = 10.86 m
the maximum height reached by the ball during its first parabolic arc is 10.86 m
b)
Rx = v₀²sin2Ф / g
Rx = 22² × sin 100 / 9.8
Rx = 48.64 m
the distance between the launch point to where the ball lands the first time is 48.64 m
c)
Rx¹ = 75% 0f Rx
Rx¹ = 0.75 × 48.64
Rx¹ = 36.48m
the distance between the launch point to where the ball lands the second time is 36.48m
Calculate the force that must be applied to an object weighing 25 pounds for it to be in equilibrium, if it is on a plane inclined at 45 °
Answer:
17.7 lb
Explanation:
Assuming the force is parallel to the incline, draw a free body diagram of the object. There are 3 forces:
Normal force N pushing perpendicular to the incline,
Weight force mg pulling down,
and applied force F pushing parallel to the incline.
Sum the forces in the parallel direction:
∑F = ma
F − mg sin 45° = 0
F = mg sin 45°
F = (25 lb) sin 45°
F = 17.7 lb
The piston of a hydraulic elevator used for lifting trucks has a 0.3m radius. What pressure is required to lift a truck of 2500 kg mass? What force was applied to the small piston if it has a radius of 3cm?
Answer:
1. 88370.45 N/[tex]m^{2}[/tex]
2. 2500 N.
Explanation:
Pressure, P, is define as the force, F, per unit area, A, applied on/ by an object.
i.e P = [tex]\frac{F}{A}[/tex]
1. The area, A, of the piston of the hydraulic elevator can be determined by;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the piston.
A = [tex]\frac{22}{7}[/tex] × [tex](0.3)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × 0.09
= 0.2829 [tex]m^{2}[/tex]
Pressure required to lift a truck of 2500 kg mass can be determined as;
P = [tex]\frac{F}{A}[/tex]
F = W = mg
= 2500 × 10
= 25 000 N
So that,
P = [tex]\frac{25000}{0.2829}[/tex]
= 88370.45 N/[tex]m^{2}[/tex]
The pressure required is 88370.45 N/[tex]m^{2}[/tex].
2. [tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Area of small piston = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × [tex](0.03)^{2}[/tex]
= 0.02829 [tex]m^{2}[/tex]
So that,
[tex]\frac{25000}{0.2829}[/tex] = [tex]\frac{F_{2} }{0.02829}[/tex]
[tex]F_{2}[/tex] = 2500 N
The force applied to the small piston is 2500 N.
A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, What was the sprinter's average velocity between the two time marks? Show all your work.
Answer:
velocity = 7.7558 m/s
Explanation:
s = u × t
s - distance u - velocity t - time
59 - 12 = (7.98 - 1.92) *u
47 = ( 6.06) u
u = 7.7558 m/s
When a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, then the sprinter's average velocity between the two-time marks would be
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. The velocity of an object depends on the magnitude as well as the direction of the object.
the mathematical expression for velocity is given by
velocity = total displacement /time
As given in the problem a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s
The total displacement covered by the sprinter
S = 59-12
S= 47 m
The total time is taken by the sprinter
T= 7.98 -1.92
= 6.06 seconds
Velocity = S/T
=47/6.06
=7.75 m/s
Thus. the average velocity of the sprinter would be 7.75 m/s
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Activity
In this activity, you will solve the following problems using the rules for adding and subtracting and multiplying and dividing with significant
figures
Part A
28.97 + 45.876
Explanation:
We need to add 28.97 and 45.876.
Here, the first no is 28.97 and other no is 45.876
Here, 28.97 is the least precise value and 45.876 is most precise.
28.97 + 45.876 = 74.846
We need to write the final answer such that there is two digit after the decimal. So, 28.97 + 45.876 = 74.85 because 6 is more than 5 i.e. we add 1 to 4.
Answer:
28.97
+ 45.876
74.846
Explanation:
Because 28.97 has only two decimal places, round off the answer to two decimal places: 74.85.
A 150 W driveway light produces 9.3 × 104 J of light energy over a 4.0 h period. If the light uses 2.06 × 106 J of energy in the 4.0 h, then what is its efficiency? Show all your work.
Answer:
4.5%
Explanation:
efficiency = energy out / energy in
e = 9.3×10⁴ J / 2.06×10⁶ J
e = 0.045
The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE/dt= q^2 a^2/ 6πεc^3, where c is the speed of light.
Required:
a. If a proton with a kinetic energy of 5.0 MeV is travelling in a particle accelerator in a circular orbit with a radius of 0.540 m, what fraction of its energy does it radiate per second?
b. Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?
Answer:
The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]
Explanation:
Given that,
Charge = q
Acceleration = a
The rate at which energy is emitted from an accelerating charge
[tex]\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon c^3}[/tex]....(I)
We know that,
Acceleration for circular motion is
[tex]a=\dfrac{v^2}{r}[/tex]....(II)
The kinetic energy is
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=\dfrac{2(K.E)}{m}[/tex]
Put the value of v in equation (II)
[tex]a=\dfrac{2(K.E)}{mr}[/tex]
Put the value of a in equation (I)
[tex]\dfrac{dE}{dt}=\dfrac{q^2(\dfrac{2(K.E)}{mr})^2}{6\pi\epsilon c^3}[/tex]
[tex]\dfrac{dE}{dt}=\dfrac{q^24(K.E)^2}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=\dfrac{q^24(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
Suppose that,
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=R[/tex]
So,
[tex]R=\dfrac{q^2\times4(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]....(III)
(a). For proton,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(1.67\times10^{-27})^2\times(0.540)^2}[/tex]
[tex]R=2.23\times10^{-11}[/tex]
(b). For electron,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(9.1\times10^{-31})^2\times(0.540)^2}[/tex]
[tex]R=0.0000753[/tex]
Hence, The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]