Answer:
tyygggghgtyhyrdfgyyyhjjillbxsrfvgygvnjj
Explanation:
cffczhxucuxoyitxohvojcdivbjv ohxc
What is the magnitude of force required to accelerate a car of mass 1.7 * 10 ^ 3 kg by 4.75 m/s
What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²
Answer:
F = 8.075 N
Explanation:
Formula for force is;
F = ma
Where;
m is mass
a is acceleration
F = 1.7 × 10³ × 4.75
F = 8.075 N
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
A marble rolling with speed 20cm/s rolls off the edge of a table that is 80cm high. How far horizontally from the table edge does the marble strike the floor
Answer:
8 cm
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 cm/s
Height (h) = 80 cm
Horizontal distance (s) =?
Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:
Height (h) = 80 cm
Acceleration due to gravity (g) = 1000 cm/s²
Time (t) =?
t = √(2h/g)
t = √[(2 × 80)/1000]
t = √(160/1000)
T = √0.16
t = 0.4 s
Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:
Initial velocity (u) = 20 cm/s
Time (t) = 0.4 s
Horizontal distance (s) =?
s = ut
s = 20 × 0.4
s = 8 cm
Thus, the horizontal distance travelled by the marble is 8 cm.
The horizontal distance traveled by the marble is 8 cm.
The given parameters;
speed of the marble, v = 20 cm/sheight of the table, h = 80 cmThe time of motion of the marble is calculated as follows;
[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s[/tex]
The horizontal distance traveled by the marble is calculated as follows;
[tex]X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm[/tex]
Thus, the horizontal distance traveled by the marble is 8 cm.
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Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
A single-turn circular loop of wire of radius 55 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 350 to 450 mT.
Required:
a. Determine the emf induced in the loop (in V). (Enter the magnitude.) V
b. If the magnetic field is directed out of the page, what is the direction of the current induced in the loop?
Answer:
Explanation:
Area of the loop = π x ( 55 x 10⁻³ )²
= 9.5 x 10⁻³ m²
Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.
time dt =.10 s
emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s
= 1.5 V .
b )
Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.
Answer:
(a) 9.5 mV
(b) clockwise
Explanation:
Radius, r = 55 mm
Time, t = 0.1 s
Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T
(a) induced emf is given by
[tex]e = A \frac{dB}{dt}[/tex]
[tex]e = A \frac{dB}{dt}\\\\e=3.14\times 0.055\times0.055\times \frac{0.1}{0.1}\\\\e= 9.5 \times 10^{-3} V = 9.5 mV[/tex]
(b) According to the Lenz law, the direction of current is clockwise.
A 0.4 m long solenoid has a total of 356 turns of wire and carries a current of 79 A. What is the magnitude of the magnetic field at the center of the solenoid?
Answer:
B = 0.088 T
Explanation:
Given that,
The length of a solenoid, l = 0.4 m
No. of turns of wire, N = 356
Current, I = 79 A
We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.
[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]
So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.
How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A) 0.50c
B) 0.65c
C) 0.78c
D) 0.87c
Answer:
Option (D) is correct.
Explanation:
Let the speed is v.
[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]
Option (D) is correct.
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
Required information
You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN
Answer:
19,224 N
Explanation:
The given parameters are;
The mass limit of the elevator = 2,400 kg
The maximum ascent speed = 18.0 m/s
The maximum descent speed = 10.0 m/s
The maximum acceleration = 1.80 m/s²
Given that the acceleration due to gravity, g ≈ 9.81 m/s²
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;
[tex]F_{min}[/tex] = m·g - m·a
∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N
If 10 W of power is supplied to 1 kg of water at 100℃, how long will it take to for the water to completely boil away? The time calculated is a little less than actual time of boiling in practice. Why?
Answer:
t = 2.26 x 10⁵ s
Explanation:
The energy supplied to the water will be equal to the heat required for the boiling of water:
E = ΔQ
Pt = mL
where,
P = Power = 10 W
t = time = ?
m = mass of water = 1 kg
L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]
t = 2.26 x 10⁵ s
This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
A stone is dropped from a bridge. It takes 4s to reach the water below. How high is the bridge above the water?
Answer:
height is 78.4m
Explanation:
h=u.t + 0.5.g.t^2
= 0 + 0.5x9.8x4^2
= 78.4m
Here we will use a simple example to demonstrate the difference between the rms speed and the average speed. Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s. Find the rms speed for this collection. Is it the same as the average speed of these molecules
Explanation:
Given that,
Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.
The rms speed for this collection is as follows :
[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]
The average speed of these molecules is :
[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]
So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.
What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?
Answer:
The velocity becomes [tex]v\sqrt 2[/tex].
Explanation:
The force acting on the bobber is centripetal force.
The centripetal force is given by
[tex]F =\frac{mv^2}{r}[/tex]
when mass remains same, radius is doubled and the force is same, so the velocity is v'.
[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]
Electric field is always perpendicular to the equipotential surface.
a. True
b. False
Answer:
a: true.
Explanation:
We can define an equipotential surface as a surface where the potential at any point of the surface is constant.
For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.
Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.
Now we want to see if the electric field is always perpendicular to these equipotential surfaces.
You can see that in the two previous examples this is true, but let's see for a general case.
Now suppose that you have a given field, and you have a test charge in one equipotential surface.
So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.
But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.
Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.
The correct option is a.
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
Why did the Prince go down on one knee?
(a) State Hook's law. [2]
(b) The walls of the tyres on a car are made of a rubber
compound. The variation with stress of the strain of a
specimen of this rubber compound is shown in Fig. 1.2.
As the car moves, the walls of the tyres end and straighten
continuously. Use Fig. 1.2 to explain why the walls of the
tyres become warm
[3]
hin
Fig. 1.2.
Hooke's law gives the relationship between the force applied and observed compression and expansion
(a) Hooke's law states that force applied is proportional to the deformation of an object
(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre
The reason for the above explanation are as follows;
(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F
Mathematically, we get;
F = k × ΔL
[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]
Given that the material cross sectional area = A, and the original length of the material = L, we get;
F/A = Stress = σ, ΔL/L = strain = ε
[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]
Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material
(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation
Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve
As the tyre straightens, the path of the stress change curve is given by the lower curve
The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening
Energy absorbed < Energy released
The balance energy is transformed into other forms of energy, including heat energy, which is observed by the raising in temperature, warming, of the tyre
Energy absorbed = Energy released + Heat energy
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Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted
between S and and q is __ the force exerted between S and p.
1/6
6 times
1/36
36 times
Answer:
Fq = k q Q / R^2
Fp = k q Q / (6 R)^2
The force exerted between S and p is 1 / 36 of that between S and q
or the force between S and q is 36 times that between S and p.
once the object is seen clearly (Figure 5).
Cliary muscles
Nea
Obec
image
25 cm
FIGURE 3
dusion : Thus, we observe that the focal length of the eye
ically by the action of cilin
Answer:
Where is the figure ?????
220V a.c. is more dangerous than 220V d.c why?
Answer:
220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.
You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 690.0 kg and was traveling eastward. Car B weighs 520.0 kg and was traveling westward at 74.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. 1) How fast (in kilometer per hour) was car A traveling just before the collision
Answer:
The speed of car A before collision is 3.5 km/h.
Explanation:
Mass of car A = 690 kg eastwards
Mass of car B = 520 kg at 74 km/h west wards
Distance, s = 6 m
coefficient of friction = 0.75
Let the speed after collision is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]
Let the initial speed of car A is v'.
Use conservation of momentum
690 x v' - 520 x 74 = (690 + 520) x 33.8
690 v' + 38480 = 40898
v' = 3.5 km/h
Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
The speed of light is the fastest in which medium
In vacuum, going at 2.99×10^8 m/s.
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m
Answer:
The gauge pressure is equal to 147 kPa.
Explanation:
The pressure exerted by fluid is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is density of water
h is height
So, put all the values,
[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]
or
P = 147 kPa
So, the gauge pressure is equal to 147 kPa.
Answer:
The gauge pressure is 147000 Pa.
Explanation:
Height, h = 15 m
density of water, d= 1000 kg/m^3
gravity, g = 9.8 m/s^2
The gauge pressure is the pressure exerted by the fluid.
The pressure exerted by the fluid is given by
P = h d g
P = 15 x 1000 x 9.8 = 147000 Pa
3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following correctly describes the tension in the three rope segments?
a. T1< T2 < T3
b. T1< T2 = T3
c. T1 = T2 = T3
d. T1> T2 > T3
please help.show how and which?
see attachment for more detail.
Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.
Let's evaluate each tension.
Case T₃.
[tex] T_{3} - W_{3} = 0 [/tex]
For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.
[tex] T_{3} = W_{3} [/tex]
Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:
[tex] T_{3} = W_{3} = mg [/tex] (1) Case T₂.
[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]
[tex] T_{2} = T_{3} + W_{2} [/tex] (2)
By entering W₂ = 2mg and equation (1) into eq (2) we have:
[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]
Case T₁.
[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]
[tex] T_{1} = T_{2} + W_{1} [/tex] (3)
Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:
[tex] T_{1} = 3mg + 3mg = 6mg [/tex]
Therefore, the correct option is d: T₁ > T₂ > T₃.
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I hope it helps you!
Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]
First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:
Mass m:
[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)
Mass 2m:
[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)
Mass 3m:
[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)
The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)
A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children plus sled is 55 kg. The mother has a mass of 61 kg. Find the static friction acting on the mother.
Answer:
f = 106.3 N
Explanation:
The force applied on the sled must be equal to the static frictional force to move the sled:
Tension Force Horizontal Component = Static Frictional Force
[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]
where,
T = Tension = 120 N
θ = angle of rope = 37°
μ = coefficient of static friction = ?
m = mass of children plus sled = 55 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]
Now, the static friction acting on the mother will be:
[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 106.3 N
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
