The mass of Car 2 was 3000 lb.
We need to apply the concept of conservation of momentum.
The velocity of both cars= 32mph
Combined velocity = 20mph
Mass of Car 1= 2000 lb
According to the conservation of momentum
M1V1+ M2V2= (M1+M2)
2000x32- (-M2x 32)=20(2000+M2)
64000+32M2=40000 +20M2
24000= 8M2
M2= 3000lb
Therefore the mass of Car 2 is 3000lb.
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A spring with spring constant 40 N/m is compressed .1m past it natural length. A mass of .5kg is attached to the spring. A. What is the elastic potential energy stored in the spring?B. The spring is released. What is the speed of the masses as it reaches the natural length of the spring?
Given data
*The given spring constant is k = 40 N/m
*The given compressed length is x = 0.1 m
*The given mass is m = 0.5 kg
(a)
The formula for the elastic potential energy stored in the spring is given as
[tex]U_p=\frac{1}{2}kx^2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(40)(0.1)^2 \\ =0.2\text{ J} \end{gathered}[/tex]Hence, the elastic potential energy stored in the spring is 0.2 J
(b)
The formula for the speed of the masses is given by the conservation of energy as
[tex]\begin{gathered} U_p=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=x\sqrt[]{\frac{k}{m}} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} v=(0.1)\sqrt[]{\frac{40}{0.5}} \\ =0.89\text{ m/s} \end{gathered}[/tex]Hence, the speed of the masses as it reaches the length of the spring is v = 0.89 m/s
A spring with a spring constant of 52N/m sits on a desk. The spring is 34cm long. A block of mass 0.12kg is placed on top of the spring. How high above the desk does the block rest?
Answer:
H = 31.7 cm
Explanation:
Given:
k = 52 N/m - Spring rate
L = 34 cm = 0.34 m - Spring length
m = 0.12 kg
g = 9.8 m/s²
____________
H - ?
Block weight:
F = m*g = 0.12·9.8 ≈ 1.18 N
According to Hooke's law:
F = k·ΔL
The spring is compressed by:
ΔL = F / k = 1.18 / 52 ≈ 0.023 м
Height of the block above the table:
H = L -ΔL = 0.34 - 0.023 = 0.317 m or H = 31.7 cm
Which part of the rock cycle can only occur because of thermal convection?
1. Magma changing to igneous rock
2. Igneous rock changing to metamorphic rock
3. Metamorphic rock changing to sedimentary rock
4. Sediment changing to sedimentary rock
Answer:
magma changing to igneous rock
Explanation:
In rock formation, there are several processes involved which could include sedimentation, melting, crystallization, etc but there is a part of the rock cycle that can only occur because of thermal convection which is magma changing to igneous rock.
This occurs when the heat from inside the earth and cooler temperature moves the rock in its liquid state which is known as convection and can also be defined as a cycle of heat transfer from where the hot material rises and the cool material condenses.
A rock is thrown off of a 120 foot cliff with an upward velocity of 20 ft/s. As a result its height after t seconds is given by the formula:h(t) = 120 + 20t - 5t^2What is its height after 2 seconds?___What is its velocity after 2 seconds?____(Positive velocity means it is on the way up, negative velocity means it is on the way down.)
We are given that the height of a rock in terms of the time is given by the following equation:
[tex]h\mleft(t\mright)=120+20t-5t^2[/tex]We are asked to determine the height after two seconds. To do that we will substitute in the equation the value of "t = 2s", like this:
[tex]h(2)=120+20(2)-5(2)^2[/tex]Solving the operations we get:
[tex]h(2)=140[/tex]Therefore, the height after 2 seconds is 140 ft.
Now, to determine an equation for the velocity we will determine the derivative with respect to the time of the equation for the height.
[tex]\frac{dh}{dt}=\frac{d}{dt}(120+20t-5t^2)[/tex]Now, we distribute the derivative:
[tex]\frac{dh}{dt}=\frac{d}{dt}(120)+\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the first derivative we will use the following rule:
[tex]\frac{d}{dt}(a)=0[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the second derivative we will use the following rule:
[tex]\frac{d}{dt}(at)=a[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=20-\frac{d}{dt}(5t^2)[/tex]For the last derivative we will use the following rule:
[tex]\frac{d}{dt}(at^n)=\text{nat}^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dh}{dt}=20-10t[/tex]Since the derivative of the position with respect to time is the velocity we have:
[tex]\frac{dh}{dt}=v=20-10t[/tex]Now, we substitute the value of "t = 2s":
[tex]v=20-10(2)[/tex]Now, we solve the operations:
[tex]\begin{gathered} v=20-20 \\ v=0 \end{gathered}[/tex]Therefore, the velocity after 2 seconds is 0.
A 0.327-kg model rocket accelerates at 35.7 m/s/s on takeoff. Determine the upward thrust experienced by the rocket
The upward thrust of the rocket is determined as 11.67 N.
What is the upward thrust of the rocket?The upward thrust of the rocket is calculated by applying Newton's second law of motion as shown below;
F = ma
where;
m is the mass of the rocketa is the upward acceleration of the rocketSubstitute the given parameters and solve for the upward thrust of the rocket.
F = (0.327 kg) x (35.7 m/s²)
F = 11.67 N
Thus, the upward thrust of the rocket is determined as 11.67 N.
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An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:
Given:
The maximum displacement from the equilibrium position is A = 0.12 m.
Half of the time period is
[tex]T_{\frac{1}{2}}=\text{ 0.8 s}[/tex]To find the amplitude, time period, and frequency.
Explanation:
Amplitude is the maximum displacement from the equilibrium position.
Thus, the amplitude is A = 0.12 m.
One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).
Thus, the time period is
[tex]\begin{gathered} T=2T_{\frac{1}{2}} \\ =2\times0.8 \\ =1.6\text{ s} \end{gathered}[/tex]The frequency will be
[tex]\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{1.6} \\ =0.625\text{ Hz} \end{gathered}[/tex]25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
Work = change in energy
Fd = 1/2 mv^2
F = 1/2 x 120 x 5^2 / 10
F = 150 N
If a student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m, then the average braking force would be 150 Newtons.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work / time
Work done by the braking force = change in kinetic energy
F × s = 1/2 × m × v²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force would be 150 Newtons.
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What does this image reveal about gravityon the moon compared to Earth?
ANSWER:
The Moon's gravity is less than Earth's.
STEP-BY-STEP EXPLANATION:
When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.
The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.
As the gravity is less, you can do things such as the one shown in the image.
As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.
What momentum does a car of mass 1,116 kg have if it’s traveling at 18m/s? Submit anwser in exponential form.
ANSWER:
20088 kg*m/s
STEP-BY-STEP EXPLANATION:
The momentum is given by the following formula:
[tex]p=m\cdot\Delta v[/tex]We know the mass and also the speed, therefore:
[tex]\begin{gathered} p=1116\cdot18 \\ p=20088\text{ kg}\cdot\frac{m}{s} \end{gathered}[/tex]Por lo tanto, el momento es igual a 20088 kg*m/s
A motorcycle has a constant acceleration of 3.74 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a)21.5 to 31.5 m/s, and (b)51.5 to 61.5 m/s?(a)Number ______ Units_________(b)Number ______ Units_________
Since the acceleration is constant we know that it is given by:
[tex]a=\frac{v_f-v_0}{t}[/tex]from where we have that:
[tex]t=\frac{v_f-v_0}{a}[/tex]once we have this equation we can determine the time by plugging the values of the acceleration and velocities.
a)
In this case we have that:
[tex]t=\frac{31.5-21.5}{3.74}=2.67[/tex]Therefore it takes 2.67 s.
b)
In this case we have that:
[tex]t=\frac{61.5-51.5}{3.74}=2.67[/tex]Therefore it takes 2.67 s.
carts, bricks, and bands
6. What acceleration results when four rubber bands stretched to 20 cm is used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
The acceleration that results when four rubber bands stretched to 20 cm is used pull a cart with one brick is about 1.00 m/s². That is option D
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
It can clearly be observed that trial 4 that made use of 4 bands resulted in an acceleration of 1.00 m/s² which is the highest observed acceleration.
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The initial earthquake had a Richter scale reading of 3.24. The aftershock was 7470 times as strong as the initial earthquake. What was the Richter scale reading of the aftershock?(a) 7.11(b) 6.48(c) 4.32(d) 3.87
In order to determine the Richter scale reading of the aftershock, use the following formula:
[tex]M_2-M_1=\log (\frac{I_2}{I_1})[/tex]Where I1 and I2 are the intensity of both earthquake and its aftershock, M1 and M2 are the magnitude on the Richter Scale for earthquake and aftershock.
In this case, you have:
I1 = intensity earthquake
I2 = 7470*I1
M1 = 3.24
M2 = ?
solve the equation above for M2 and replace the values of the given parameters:
[tex]undefined[/tex]
A light, inextensible cord passes over alight, frictionless pulley with a radius of15 cm. It has a(n) 18 kg mass on the left and a(n) 2.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1.5 m apart.The acceleration of gravity is 9.8 m/s².
At what rate are the two masses accelerating when they pass each other answer in units of m/s^2
Answer:
quizlet
Explanation:
they help
Consider a fluid of density 3.43 g⋅cm−3 flowing through a pipe of varying cross-section. The diameter of the pipe in one section is 9.1 cm, while the diameter in a second section is 12.6 cm. When the diameter of the pipe is 9.1 cm, the flow speed of the fluid is 339 cm⋅s−1 and the pressure is 2.93 × 105 Pa.A)Calculate the flow speed (in m⋅s−1) of the fluid when the diameter of the pipe is 12.6 cm. B)Calculate the pressure (in × 105 Pa) when the pipe has a diameter of 12.6 cm
Given that the pipe has varying cross-sections.
The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.
Also, the fluid has the density,
[tex]\rho=3.43gcm^{-3}[/tex]The area of the cross-section for the first section is
[tex]\begin{gathered} A_1=\frac{\pi(d1)^2}{4} \\ =\frac{\pi(9.1)^2}{4}cm^2 \end{gathered}[/tex]The area of the cross-section for the second section is
[tex]\begin{gathered} A_2=\frac{\pi(d2)^2}{4} \\ =\frac{\pi(12.6)^2}{4}cm^2 \end{gathered}[/tex]The flow speed for the first section is v1 = 339 cm s^-1
The flow speed for the second section will be v2.
(a) The flow speed for the second section can be calculated as
[tex]\begin{gathered} A_1v1=A_2_{}v2 \\ v2=\frac{A_1v1}{A_2} \\ =\frac{\pi(9.1)^2\times339\times4}{4\times\pi\times(12.6)^2} \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}[/tex](b) The pressure for first section is p1 = 2.93 x 10^5 Pa
The pressure for the second section will be p2.
The pressure for the second section can be calculated by the formula,
[tex]\begin{gathered} p2=p1+\frac{1}{2}\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93\times10^5+\frac{1}{2}\times3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }\times10^5\text{ Pa} \end{gathered}[/tex]If there is a source voltage of 12 volts with a 2.5 volt drop over the LED, and an intended current of 100 mA, what size resistor should be used?
Given:
A voltage source has a voltage of 12 V.
A voltage drop over the LED is 2.5 V.
The current in the circuit is 100 mA.
To find:
The resistance of the resistor should be used.
Explanation:
LED is a forward-biased PN junction that emits light. The voltage drop over the LED is 2.5 V.
The voltage source has a voltage of 12 V.
Hence, the total voltage in the circuit = 12 V - 2.5 V = 9.5 V
The resistance of the circuit in the forward biased is almost equal to zero. Let the resistance in the circuit be R.
Thus, the total resistance in the circuit = 0 + R = R
The current in the circuit is = 100 mA = 0.1 A
The resistance R of the circuit can be calculated as:
[tex]R=\frac{V}{I}[/tex]Here, V is the voltage in the circuit and I is the current in the circuit.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} R=\frac{9.5\text{ V}}{0.1\text{ A}} \\ \\ R=95\text{ }\Omega \end{gathered}[/tex]Final answer:
Hence, a resistor of 95 Ω should be used in the circuit.
Three resistors having values of 4 Ω, 6 Ω , and 8Ω are connected in series. Their equivalent resistance is ______.Group of answer choices18 Ω8 Ω6 Ω1.80 Ω
Answer:
18Ω
Explanation:
If the resistors are connected in series, the equivalent resistance is the sum of each resistance, so
Equivalent resistance = 4Ω + 6Ω + 8Ω
Equivalent resisteance = 18Ω
Therefore, the answer is 18Ω
Alpha decay results in the emission of aPositively charged helium ionPhotonNegatively charged electronNeutrino
An alpha particle is a Helium nucleus that is emitted during alpha decay, this meant that alpha decay results in the amission of a Positively charged helium ion.
7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m
The final answer is
(a) The KE = 3.88 * 10
(b) The speed of the arrow is 2.93 * 10
(c) Height which the arrow attains is 4.38 * 10
When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.
Given,
Work done = 7 J
Mass of the arrow = 0.09 kg
The computations for a and b are the following:
To calculate the Kinetic energy,Work done = force x distance
7.0 J = force x 0.09
force = 7/0.09=77.77 = 7.77 x 10
force = m. a = 0.09 x a
77.77 / 0.09 = a = 864.1 = 8.64*10²
KE = 0.5 x m x v²
= 0.5 x 0.09 x 864.1
= 3.88 * 10
The speed of the arrow can be calculated asspeed 38.8/(0.5 x 0.09) = v² = 862 sq-root
Hence, v = 2.93*10¹ m/s is the answer
Height that the arrow reaches29.3² = 2gh
= 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m
h = 43.8 m or 4.38 * 10¹ is the answer
Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.
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Instructions: 1) write out the question 2) work out the solution 3) Explain in words how you would know to do that1. What is the momentum of a 5.0 g bullet with a velocity of 500 m/s?
Given data:
The mass of bullet is m=5.0 g.
The velocity fo bullet is v=500 m/s.
The formula for the momentum is given by,
[tex]p=mv[/tex]Substitute the given values in above equation,
[tex]\begin{gathered} p=(5g\times\frac{1kg}{1000\text{ g}})(\frac{500m}{s}) \\ p=\frac{2.5kgm}{s} \end{gathered}[/tex]Thus, the momentum of the bullet is 2.5 kgm/s.
According to the definition of momentum, "It is the product of mass and velocity". Therefore, the momentum of moving body can be calculated by multiplying the mass of the body and velocity at which the body is moving.
So, that is how we know how to calculate the momentum of the body.
A hydroelectric plant takes energy from water and turns it into electrical energy.What are the transformations of energy in the water molecules that are used in theprocess of generating electricity this way?The water particles initially have kinetic energy due to their motion. This kineticenergy is transformed into potential energy due to the position of the water andthen this energy is used to produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into thermal energy and then this energy is used toproduce electricity.The water particles initially have chemical energy due to the bonds in water. Thischemical energy is transformed into thermal and then the thermal energy is usedto produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into kinetic energy due to the motion of energy and thenthe kinetic energy is used to produce electricity.
Answer:
The last option.
Explanation:
The principle of hydroelectric dams is the following:
Basically, we utilizie the energy of the falling wat
Harry Hand can run 100m in 20s. His daughter, Linda Hand, canrun 50m in 8.5s. Who was faster
The speed is given as;
[tex]v=\frac{d}{t}[/tex]Here, d is the distance covered and t is the time.
Harry Hand covers a distance of 100 m (d_H) in 20 s (t_H). Therefore, the speed of the Harry Hand is,
[tex]v_H=\frac{d_H}{t_H}[/tex]Substituting all known values,
[tex]\begin{gathered} v_H=\frac{100\text{ m}}{20\text{ s}} \\ =5.0\text{ m/s} \end{gathered}[/tex]Now, Linda Hand covers a distance of 50 m (d_L) in 8.5 s (t_L). Therefore, the speed of Linda Hand is given as,
[tex]v_L=\frac{d_L}{t_L}[/tex]Substituting all known values,
[tex]\begin{gathered} v_L=\frac{50\text{ m}}{8.5\text{ s}} \\ \approx5.88\text{ m/s} \end{gathered}[/tex]Since the speed of Linda Hand is greater than Harry Hand (v_L>v_H). Therefore, Linda Hand is faster.
An electron is in an infinite one dimension well that is 8.9 nm wide. What is the ground state energy of the electron?
The ground state energy of the electron is 2.23 x 10⁻¹⁷ J.
What is the ground state energy of the electron?The ground state energy of the electron is calculated by applying the formula for energy of photons.
E = hf
E = hc/λ
where;
h is Planck's constantc is speed of lightλ is the wavelengthE = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (8.9 x 10⁻⁹)
E = 2.23 x 10⁻¹⁷ J
Thus, the ground state energy of the electron is determined by applying the principle or formula for energy of a single photon at the given wavelength of 8.9 nm.
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jerome pitches a baseball of mass 0.300 kg. the ball arrives at home plate with a speed of 40.0 m/s and is batted straight back to jerome with a return speed of 52.0 m/s. what is the magnitude of change in the ball's momentum?
The change in the momentum of the ball is 3.6 Kgm/s.
What is momentum?The term momentum has to do with the product in the velocity of a body and mass of the body. We have to recall at this point that rate of change of momentum is directly related to the impressed force and this is in accordance with the Newton second law of motion.
Now we have to look at the few pieces of information that we can be able to glean from the question;
Mass of the object = 0.300 kg
Initial speed of the object = 40.0 m/s
Final speed of the object = 52.0 m/s
Given that the change in the velocity of the object is given by;
m( v - u)
m = Mass of the object
v = Final speed of the object
u = Initial speed of the object
change in the velocity of the object = 0.300(52 - 40)
= 3.6 Kgm/s
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Describe
protons.
Location:
Charge:
Mass:
The protons in an atom are classified according to their mass, charge, and location as follows:
Particle: Protons
Mass: 1.67262 × 10⁻²⁷ kg
Charge: Positive charge (+e or +1)
Location: Found in the nucleus of every atom
The proton is a stable subatomic particle with a rest mass of 1.67262 x 10⁻²⁷ kg, or 1,836 times the mass of an electron, with a positive charge that is equal to one electron's charge in magnitude.
All atomic nuclei, with the exception of the hydrogen nucleus, are composed of protons and neutrons, which are electrically neutral particles (that consist of a single proton). A given chemical element's nuclei all contain the same number of protons. This number establishes an element's atomic number and establishes the element's position in the periodic table. An atom is electrically neutral when the number of protons in its nucleus equals the number of electrons in its orbit.
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Every time a cycle of convection currents occur, the temperature of the water ________.A. IncreasesB. Stays the sameC. DecreasesD. Goes to zero
The answer to this question is B
The reason the answer is B is because of how the cycle works. In the beggining, water is heated up and evaporates into the atmosphere as water vapor. Then the water cools down and condensates, coming back down. The water then ends up back to the same temperature as the earth. So in the end, the entire cycle doesn't add or remove heat from the water.
A Curve in a road has 60m radius The angle of bank of the road is 4.7° Find the maximum speed of car can have without skidding. If the Coefficient of Static Friction between tyres and road is 0.8.
Answer:
An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is 0.75, is.
Explanation:
Calculate the kinetic energy of a 3.15 kg bowling ball rolling down the lane at 2.5 m/s. Include the units.
Given:
The mass of the ball is: m = 3.15 kg
The velocity of the ball is: v = 2.5 m/s
To find:
The kinetic energy of the ball.
Explanation:
The kinetic energy is the energy that a particle has when is is in motion. For the particle of mass m moving with velocity v, the kinetic energy KE is given as:
[tex]KE=\frac{1}{2}mv^2[/tex]Substitute the values in the above expression, we get:
[tex]\begin{gathered} KE=\frac{1}{2}\times3.15\text{ kg}\times(2.5\text{ m/s\rparen}^2 \\ \\ KE=\frac{1}{2}\times3.15\text{ kg}\times6.25\text{ m}^2\text{/s}^2 \\ \\ KE=9.84\text{ kg.m}^2\text{/s}^2 \\ \\ KE=9.84\text{ J} \end{gathered}[/tex]Final answer:
The kinetic energy of the ball is 9.84 J.
Which of the following is needed for an electric circuit to work?A. Fusion sourceB. Nuclear sourceC. Chemical sourceD. Energy sourceVoltage is a measure of the ________ it gives to the current.A. AccelerationB. SpeedC. PullD. Push
We will have the following:
The "part" needed for an electric circuit is an energy source.
Voltage is the measure of the acceleration it gives to the current.
A negative charge of .30 c and a positive charge of .50 are separated by .40What is the force between the charges?
Given:
The negative charge q1 = 0.3 C
The positive charge q2 = 0.5 C
The distance between the charges is 0.4 m
To find the magnitude of the force between them.
Explanation:
The formula to calculate the magnitude of the force is
[tex]F=\frac{kq1q2}{r^2}[/tex]Here, k is Coulomb's constant whose value is
[tex]k=\text{ 9}\times10^9Nm^2C^{-2}[/tex]On substituting the values, the magnitude of force will be
[tex]\begin{gathered} F=\frac{9\times10^9\times0.3\times0.5}{(0.4)^2} \\ =8.4375\text{ }\times10^9\text{ N} \end{gathered}[/tex]Final Answer: The magnitude of the force is 8.4375 x 10^(9) N.
What do you diagram to analyze orbital motion ?
The diagram to analyze the orbital motion can be shown as,
Here, a is the acceleration of moon and v is the speed.
The above diagram indicates the orbital motion of the moon around the earth. The moon is more towards the earth than the sun due to larger gravity of earth and at the same time the moon has its velocity that tends moon to move. Therefore, the moon has balanced gravitational and centripetal force to keep in an uniform orbital motion.