Answers
Explanation:
Below is an attachment containing the solution.
To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
Explanation:To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electric force is:
F = k * (q1 * q2) / r^2
Where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the charge of the Moon is 27.3% of the charge of the Earth, and the radius of the Moon is also 27.3% of the radius of the Earth. Using these values, we can calculate the electric force.
Let's assume the charge of the Earth is q1 and the charge of the Moon is q2. Since the charge of the Moon is 27.3% as large as the charge of the Earth, we can write q2 = 0.273 * q1. Similarly, the radius of the Moon is 27.3% of the radius of the Earth, so we can write r = 0.273 * R, where R is the radius of the Earth. Plugging these values into Coulomb's law formula:
F = k * (q1 * (0.273 * q1)) / (0.273 * R)^2
Simplifying the equation, we get:
F = k * (q1^2 * 0.273) / (0.273^2 * R^2)
The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
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Related Questions
Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer I0 and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of φ = 2θ.
Answers
Answer:
Explanation:
Intensity of unpolarised light = I₀
intensity after passing through first polariser = I₀ / 2
Angle between first and second polariser is φ so
intensity after passing through second polariser
= (I₀ / 2) cos²φ
Now angle between second and third polariser
= 90 - φ
intensity after passing though third polariser
= (I₀ / 2) cos²φ cos²( 90 - φ)
= (I₀ / 2) cos²φ sin²φ
= (I₀ / 8) 4cos²φ sin²φ
= (I₀ / 8) sin²2φ
The intensity after the third polarizer will be:
I₃ = (I₀/8)*sin^2(2φ)
What is the resulting intensity?
For non-polarized light that passes through any polarizer, we say that the intensity is reduced to its half.
Original intensity = I₀
After the first polarizer, the intensity will be:
I₁ = I₀/2.
Now, when it passes through a polarizer such that the difference in angles with the polarization is x, the new intensity will be:
I₂ = I₁*cos^2(x).
The angle between the second and the first polarizer is φ, then we have:
I₂ = (I₀/2)*cos^2(φ).
Now we also know that the first and the last polarizer are crossed (so there is an angle of 90°). Then if we define θ as the angle between the second and the third polarizer, we will have that:
φ + θ = 90°
then:
θ = 90° - φ
The intensity after the third polarizer will be:
I₃ = (I₀/2)*cos^2(φ)*cos^2(90° - φ)
And we know that:
cos(90° - φ) = sin(φ)
Then we can rewrite:
I₃ = (I₀/2)*cos^2(φ)*sin^2(φ)
But we want a single trigonometric function, then we use the relation:
cos^2(φ)*sin^2(φ) = sin^2(2φ)/4
And replacing that, we get:
I₃ = (I₀/2)*sin^2(2φ)/4 = (I₀/8)*sin^2(2φ)
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If the electric field has a magnitude of 460 N/C and the magnetic field has a magnitude of 0.16 T, what speed must the particles have to pass through the selector undeflected?
Answers
The speed of the particle to pass through the Selector is 2875 m/s
Explanation:
Given -
Electric field,(We can represent as E) = 460 N/C
Magnetic field, (We can represent as B) = 0.16 T
Speed,(We can represent as v) = ?
We know that the formula for finding the velocity ,
[tex]v = \frac{E}{B} \\\\v = \frac{460}{0.16} \\\\v = 2875m/s[/tex]
Therefore, speed of the particle to pass through the Selector is 2875 m/s
In pushing a heavy box across the floor, is the force you need to apply to start the box moving greater than, less than, or the same as the force needed to keep the box moving? On what are you basing your choice?
How do you think the force of friction is related to the weight of the box? Explain.
Answers
Answer:
Explanation:
Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .
A threshold force is needed to move the box and when box started to move kinetic friction comes into play.
Friction force is directly related to the weight of the box as the friction force is
coefficient of friction time Normal reaction .
And Normal reaction is equal to the weight of box if no force is applied.
[tex]f_r=\mu N[/tex]
[tex]N=mg[/tex]
The force to start moving a box is greater than to keep it moving due to static and kinetic friction. The force of friction is directly proportional to the weight of the box; the heavier the box, the greater the friction.
Explanation:In physics, the force required to start moving an object is often greater than the force required to keep it moving. This is due to a concept known as friction, particularly, static friction and kinetic friction. Static friction is the force that resists the initiation of sliding motion, and it's usually greater than kinetic friction, which is the force that opposes motion of an object when the object is in motion.
Now, relating friction to the weight of the box, the weight of an object is equal to its mass times the acceleration due to gravity, and it's directly proportional to the force of friction. In essence, the heavier the box is, the more the static and kinetic friction that you have to overcome to move and keep it moving.
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The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
Answers
Question is not complete and the missing part is;
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
Answer:
0.828 m/s
Explanation:
Resolving vertically, we have;
Fn and Fg act vertically. Thus,
Fn - Fg = 0 - - - - eq(1)
Resolving horizontally, we have;
Ff = ma - - - - eq(2)
Now, Fn and Fg are both mg and both will cancel out in eq 1.
Leaving us with eq 2.
So, Ff = ma
Now, Frictional force: Ff = μmg where μ is coefficient of friction.
Also, a = v²/r
Where v is linear speed or velocity
Thus,
μmg = mv²/r
m will cancel out,
Thus, μg = v²/r
Making v the subject;
rμg = v²
v = √rμg
Plugging in the relevant values,
v = √0.14 x 0.5 x 9.8
v = √0.686
v = 0.828 m/s
Final answer:
To determine the linear speed of the coin when it just begins to slip, we can use the equation frictional force = centripetal force for circular motion. By equating these two forces and solving for the linear speed, we can find the answer.
Explanation:
To determine the linear speed of the coin when it just begins to slip, we can use the equation:
frictional force = centripetal force for circular motion
The frictional force can be calculated using the equation:
frictional force = coefficient of static friction x normal force
And the centripetal force can be calculated using the equation:
centripetal force = mass of coin x acceleration towards the center of the disk
By equating these two forces and solving for the linear speed, we can find the answer.
In this case, we are given the coefficient of static friction as 0.50. We can also assume that the normal force is equal to the weight of the coin, which is the mass of the coin multiplied by the acceleration due to gravity. By plugging in these values, we can find the linear speed of the coin.
A factory worker moves a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.
1. What magnitude of force must the worker apply?
2. How much work is done on the crate by the worker's push?
3. How much work is done on the crate by friction?
4. How much work is done by normal force? By gravity?
5. What is the net work done on the crate?
Answers
To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,
PART A) The friction force with the given data is,
[tex]F_f = \mu_k mg[/tex]
Here,
[tex]\mu_k[/tex] = Kinetic coefficient
m = Mass
g = Gravitational acceleration
[tex]F_f = (0.25)(30kg)(9.8m/s^2)[/tex]
[tex]F_f = 73.5N[/tex]
PART B) The work done by the worker is the distance traveled for the previously force found, then
[tex]W_w = rF_f[/tex]
[tex]W_w = (4.5m)(73.5N)[/tex]
[tex]W_w = 330.75J[/tex]
PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore
[tex]W_f = -rF_f[/tex]
[tex]W_f = -(4.5m)(73.5N)[/tex]
[tex]W_f = -330.75J[/tex]
[tex]W_f = -331J[/tex]
PART D) The work done by the normal force is,
[tex]W_N = N (r) Cos(90)[/tex]
[tex]W_N = 0J[/tex]
The work done by gravitational force is
[tex]W_g = rF_gcos(90)[/tex]
[tex]W_g = 0J[/tex]
PART E) The expression for the total work done is,
[tex]W_{net} = W_f +W_w +W_N+W_g[/tex]
[tex]W_{net} = 330.75-330.75+0+0[/tex]
[tex]W_{net} = 0J[/tex]
Therefore the net work done by the system is 0J
Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.
Answers
To calculate the distance L between the screen and the slit, use the single-slit diffraction minimum condition combined with the known distance of the red laser's third minimum and its wavelength.
Explanation:The student is working on a Physics problem related to single-slit diffraction. To find the distance L between the screen and the slit, we can use the formula for the position of a diffraction minimum, y = L × tan(θ), where θ is the angle of the diffraction minimum. However, for small angles (as in most diffraction problems), tan(θ) ≈ sin(θ), so the formula simplifies to y = L × sin(θ). The condition for the minima in a single-slit diffraction pattern is given by a × sin(θ) = m × λ, where a is the width of the slit, m is the order number of the minimum, and λ is the wavelength of the light. Given that the third diffraction minimum (m = 3) for the red laser is at a distance y3,red = 4.05 cm from the pattern's center, and the wavelength of the red light is λ = 633.0 nm, we can write 0.150 mm × sin(θ) = 3 × 633.0 nm. Solving for sin(θ) and then using y = L × sin(θ) where y = 4.05 cm, we can calculate the distance L.
When switch S is open, the voltmeter across the battery reads 1.52V. When theswitch is closed , the voltmeter reading drops to 1.37V and the ammeter reads 1.5A.Find the internal resistance of the battery.
Answers
Answer:
r = 0.1 Ω
Explanation:
We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.
When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.
[tex]0.15 = (1.5)r\\r = 0.1~\Omega[/tex]
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.599 times that of the initial number of microstates?
Answers
Answer:
Entropy Change = 0.559 Times
Explanation:
Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 cm2 . If a voltage of 145 V is applied to the arrangement, determine the voltage across the 2.5 cm2 wire.
Answers
Answer:
22.1 V
Explanation:
We are given that
[tex]A_1=0.7 cm^2=0.7\times 10^{-4} m^2[/tex]
[tex]A_2=2.5 cm^2=2.5\times 10^{-4} m^2[/tex]
[tex]A_3=2.2 cm^2=2.2\times 10^{-4} m^2[/tex]
[tex]A_4=3 cm^2=3\times 10^{-4} m^2[/tex]
Using [tex] 1cm^2=10^{-4} m^2[/tex]
We know that
[tex]R=\frac{\rho l}{A}[/tex]
In series
[tex]R=R_1+R_2+R_3+R_4[/tex]
[tex]R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}[/tex]
[tex]R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}[/tex]
Substitute the values
[tex]R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})[/tex]
[tex]R=\rho l(2.62\times 10^4)[/tex]
[tex]V=145 V[/tex]
[tex]I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}[/tex]
Voltage across the 2.5 square cm wire=[tex]IR=I\times \frac{\rho l}{A_2}[/tex]
Voltage across the 2.5 square cm wire=[tex]\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V[/tex]
Voltage across the 2.5 square cm wire=22.1 V
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant a=0.352 nm. Calculate the density of Iron β Round your answer to 3 significant digits. cm
Answers
Answer:
8.60 g/cm³
Explanation:
In the lattice structure of iron, there are two atoms per unit cell. So:
[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.
Therefore:
[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]
This implies that:
[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]
= [tex]\frac{4}{\sqrt{3} }r[/tex]
Assuming that there is no phase change gives:
[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]
= 8.60 g/m³
. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C and mp = 1.67 · 10-27 kg)
Answers
Answer:
[tex]9.198\times 10^6 m/s[/tex]
Explanation:
We are given that
Magnetic field, B=1.2 T
Radius of circular path, r=0.080 m
[tex]q_p=1.6\times 10^{-19} C[/tex]
[tex]m_p=1.67\times 10^{-27} kg[/tex]
[tex]\theta=90^{\circ}[/tex]
We have to find the speed of proton.
We know that
Magnetic force, F=[tex]qvBsin\theta[/tex]
According to question
Magnetic force=Centripetal force
[tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]
[tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]
[tex]v=9.198\times 10^6 m/s[/tex]
Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm
Answers
Answer:
0.0399 m
Explanation:
We are given that
Mass of coffee=375g=[tex]\frac{375}{1000}=0.375 kg[/tex]
1kg=1000g
Depth=h=7.5 cm=[tex]7.5\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
Density of coffee=[tex]\rho=1000kg/m^3[/tex]
We have to find the inside radius of coffee mug.
We know that
[tex]\rho=\frac{m}{V}[/tex]
Substitute the values
[tex]1000=\frac{0.375}{\pi r^2h}[/tex]
[tex]r^2=\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}[/tex]
By using [tex]\pi=3.14[/tex]
[tex]r=\sqrt{\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}}[/tex]
[tex]r=0.0399 m[/tex]
Hence, the inside radius=0.0399 m
A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 42 m/s. The third piece has three times the mass of the other two pieces. Randomized Variables V = 42 m/s show answer No Attempt 33% Part (a) Write an expression for the speed of the larger piece, that is in terms of only the variable V.
Answers
Answer:
Speed of larger piece is [tex]\dfrac{V\sqrt{2}}{3}[/tex]
Explanation:
We apply the principle of conservation of momentum.
The watermelon is initially at rest. The initial momentum = 0 kg m/s in all directions.
After the collision,
Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = [tex]mV + 3mv_{ly}[/tex]
Here, [tex]v_{ly}[/tex] is the y-component of velocity of larger piece.
This is equal to 0, since the initial momentum is 0.
[tex]v_{ly}=\dfrac{V}{3}[/tex]
Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = [tex]mV + 3mv_{lx}[/tex]
Here, [tex]v_{lx}[/tex] is the x-component of velocity of larger piece.
This is also equal to 0, since the initial momentum is 0.
[tex]v_{lx}=\dfrac{V}{3}[/tex]
The velocity of the larger piece, [tex]v_l[/tex], is the resultant of [tex]v_{lx}[/tex] and [tex]v_{ly}[/tex]. Since they are mutually perpendicular,
[tex]v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}[/tex]
[tex]v_l = \dfrac{V\sqrt{2}}{3}[/tex]
A power plant burns 1000 kg of coal each hour and produces 500 kW of power. Calculate the overall thermal efficiency if each kg of coal produces 6 MJ of energy.
Answers
Answer:
The overall thermal efficiency is 30%.
Explanation:
Given;
Output power = 500 kWh
input energy per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh
1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh
Thus, total input power = 1666.67 kWh
Overall thermal efficiency = Total output power/Total input Power
Overall thermal efficiency = (500/1666.67) *100
Overall thermal efficiency = 0.29999 *100
Overall thermal efficiency = 30%
Therefore, the overall thermal efficiency is 30%.
The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10−18 JJ of kinetic energy in this distance?
Answers
The solution is in the attachment
Answer:
[tex]a=1.1*10^{18}\frac{m}{s^2}[/tex]
Explanation:
We use the following kinematic formula to calculate the acceleration:
[tex]v_f^2=v_0^2+2ax[/tex]
The kinetic energy is defined as:
[tex]\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}[/tex]
Replacing this in the acceleration formula and solving for a:
[tex]\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}[/tex]
A constant force N acts on an object as it moves along a straight-line path. If the object’s displacement is m, what is the work done by this given force?
Answers
Answer:
Work = N x m
Explanation:
W=Fxd
Work equals force times the distance (displacement).
A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it has traveled 4.20 m downward
Answers
The speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.
Explanation:To calculate the speed of the ball after it has traveled 4.20 m downward, we need to use the principles of free fall and the equations of motion. Since the ball is dropped from a height of 55.0 m, we can calculate the initial velocity using the equation v_i = sqrt(2 * g * h), where v_i is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (55.0 m). Plugging in these values, we find that the initial velocity is approximately 34.02 m/s.
Next, we can calculate the final velocity using the equation v_f = sqrt(v_i^2 + 2 * g * d), where v_f is the final velocity, v_i is the initial velocity, g is the acceleration due to gravity, and d is the distance traveled downward (4.20 m). Plugging in the values, we get v_f = sqrt((34.02 m/s)^2 + 2 * (9.8 m/s^2) * (4.20 m)) = approximately 24.04 m/s.
Therefore, the speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.50 x 10-4 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry? Express your answer using two significant figures.
Answers
Answer:
[tex]I = 37.5\ A[/tex]
Explanation:
Given,
Magnetic field,B = 0.5 x 10⁻⁴ T
distance,r= 15 cm = 0.15 m
Current = ?
Using Ampere's law of magnetic field
[tex]B = \dfrac{\mu_0I}{2\pi r}[/tex]
[tex]I= \dfrac{B (2\pi r)}{\mu_0}[/tex]
[tex]I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}[/tex]
[tex]I = 37.5\ A[/tex]
Current in the wire is equal to [tex]I = 37.5\ A[/tex]
The maximum current this wire can carry is equal to 37.5 Amperes.
Given the following data:
Magnetic field = [tex]0.5 \times 10^{-4}\;T[/tex].Distance = 15 cm to m = 0.15 meter.Scientific data:
Permeability of free space = [tex]4\pi \times 10^{-7}[/tex]How to calculate the maximum current.In order to determine the maximum current, we would apply Ampere's law of magnetic field.
Mathematically, Ampere's law of magnetic field is given by this formula:
[tex]I=\frac{2B\pi r}{\mu_o }[/tex]
Where:
B is the magnetic field.I is the current.r is the distance.[tex]\mu_o[/tex] is the permeability of free space.Substituting the given parameters into the formula, we have;
[tex]I=\frac{2 \pi \times 0.5 \times 10^{-4}\times 0.15}{4\pi \times 10^{-7} }\\\\I=\frac{0.5 \times 10^{-4}\times 0.15}{2\pi \times 10^{-7} }[/tex]
I = 37.5 Amperes.
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A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoidally with time. With the positive x-axis pointed towards the East, the x-component of the force is given as follows:
Fx = F₀cos(ωt), where F₀ = 2 N and ω = 1.1 rad/s.
At t = 0 the particle is at x₀ = 0 and has the x-component of the velocity, vₓ = 0.
What is the x-component of velocity (vₓ) in meters per second at t= 1.5 seconds?
Answers
Answer:
[tex]v(1.5)=0.7648\ m/s[/tex]
Explanation:
Dynamics
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
[tex]\displaystyle a=\frac{F}{m}[/tex]
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
[tex]F=2cos1.1t[/tex]
The variable acceleration is calculated by:
[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]
[tex]a=0.8439cos1.1t[/tex]
The instant velocity is the integral of the acceleration:
[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]
[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]
Integrating
[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]
[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]
[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]
Final answer:
To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.
Explanation:
The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.
Two speakers are 3.0 m apart and play identical tones of frequency 170 Hz. Sam stands directly in front of one speaker at a distance of 4.0 m. Is this a loud spot or a quiet spot? Assume that the speed of sound in air is 340 m/s.
Answers
Answer:
he phase difference is π the destructive interference and the lujar is a still or silent place
Explanation:
This is a sound interference exercise where the amino difference is equal to the phase difference of the sound.
Δr / λ = ΔФ / 2π
Let's find the path difference
r₁ = 4m
r₂ = √ (4² + 3²) = 5 m
Δr = r₂ - r₁
Δr = 5-4 = 1m
Let's find the wavelength of the sound
v = λ f
λ = v / f
λ = 340/170
λ = 2m
Let's find the phase difference between the two waves
ΔФ = Δr 2π / λ
ΔФ = 1 2π / 2
ΔФ = π
Since the phase difference is π the destructive interference and the lujar is a still or silent place
Final answer:
To determine if Sam is standing on a loud or quiet spot, the path difference of the sound waves from the speakers was calculated, which is half the wavelength, indicating that Sam stands at a loud spot due to constructive interference.
Explanation:
The question relates to interference patterns created by the sound waves from two speakers and whether the spot where Sam stands is a loud or quiet spot. To determine this, we need to calculate the path difference of the sound waves reaching Sam from both speakers. The speed of sound in air is given as 340 m/s, and the frequency of the tone is 170 Hz.
The wavelength (λ) of the sound can be found using the formula speed = frequency × wavelength, which results in λ = 340 m/s / 170 Hz = 2 m. The path difference is the difference in distance from each speaker to Sam. For one speaker, the distance is 4.0 m. For the other speaker, we use the Pythagorean theorem since Sam is standing in front of the first speaker: √(4.0^2 + 3.0^2) = 5.0 m. The path difference is therefore 5.0 m - 4.0 m = 1.0 m, which is exactly half the wavelength.
Since the path difference corresponds to half a wavelength, this results in constructive interference, and Sam is indeed standing at a loud spot.
A 2.00 g air‑inflated balloon is given an excess negative charge, q 1 = − 3.75 × 10 − 8 C, by rubbing it with a blanket. It is found that a charged rod can be held above the balloon at a distance of d = 6.00 cm to make the balloon float. Assume the balloon and rod to be point charges. The Coulomb force constant is 1 / ( 4 π ϵ 0 ) = 8.99 × 10 9 N ⋅ m 2 / C 2 and the acceleration due to gravity is g = 9.81 m / s 2 .
Answers
Answer:
(+2.093 × 10⁻⁷) C
Explanation:
Coulomb's law gives the force of attraction between two charges and it is given by
F = kq₁q₂/r²
where q₁ = charge on one of the two particles under consideration = charge on the balloon = - 3.75 × 10⁻⁸ C
q₂ = charge on the other body = charge on the rod = ?
k = Coulomb's constant = 1/(4 π ϵ₀) = 8.99 × 10⁹ N⋅m²/C²
r = distance between the two charges = d = 6.00 cm = 0.06 m
But for this question, the force of attraction between the charges was enough to lift the balloon and match its weight, Hence,
F = (kq₁q₂/d²) = - mg (negative because it's an attractive force)
m = mass of balloon = 2.00 g = 0.002 kg
g = acceleration due to gravity = 9.8 m/s²
(8.99 × 10⁹ × (-3.75 × 10⁻⁸) × q₂)/(0.06²) = 0.002 × 9.8
q₂ = (-0.002 × 9.8 × 0.06²)/(8.99 × 10⁹ × (-3.75 × 10⁻⁸)
q₂ = + 2.093 × 10⁻⁷ C
The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker. Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer
Answers
Answer: 0.321 seconds
Explanation:
Let assume that air has a temperature of 20 °C. Sound speed at given temperature is [tex]343 \frac{m}{s}[/tex]. As sound spreads at constant speed, time can be easily found by using this formula:
[tex]\Delta t = \Delta t_{far} - \Delta t_{near}[/tex]
[tex]\Delta t = \frac{x_{far}-x_{near}}{v_{sound,air}}[/tex]
[tex]\Delta{t} = \frac{120 m - 10 m}{343 \frac{m}{s} }\\\\\Delta {t} = 0.321 sec[/tex]
A major artery with a cross-sectional area of 1.00cm2 branches into 18 smaller arteries, each with an average cross-sectional area of 0.400cm2. By what factor is the average velocity of the blood reduced when it passes into these branches
Answers
Answer:
The velocity in the smaller arteries will be reduced by a factor of 0.139
Explanation:
The flow rate of blood is going to stay the same when it is transferred from the major artery to the smaller ones.
flow rate = Velocity * Area
Since the flow rate remains constant, we have:
Flow rate in major artery = combined flow rate in smaller arteries
Velocity in Major artery * 1.00 = Velocity in smaller artery * (0.4 * 18)
[tex]V_M * 1 = V_S * (18*0.4)[/tex]
[tex]\frac{V_S}{V_M}=\frac{1}{18*0.4}[/tex]
[tex]\frac{V_S}{V_M}= 0.139[/tex]
Thus, the velocity in the smaller arteries will be reduced by a factor of 0.139
When electric power plants return used water to a stream, after using it in their steam turbines and condensers, this used water can lead to ____________ pollution.
Answers
Answer:
Thermal Pollution
Explanation:
Thermal pollution is a term use to describe the final result caused from the water used as a coolant by power plants and/or industrial companies. It changes the water temperature and its quality. It is usually caused from humans or companies especially manufacturing companies - They use it for their personal needs. The end product of the water after being discarded is Thermal Pollution
You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.23 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is mg, the weight of the car, where as usual g = +9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.23 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.
(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed =_______________ m/s
Answers
Answer:
[tex]v=12.65\ m.s^{-1}[/tex]
Explanation:
Given:
mass of vehicle, [tex]m=1350\ kg[/tex]radius of curvature, [tex]r=71\ m[/tex]coefficient of friction, [tex]\mu=0.23[/tex]During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:
[tex]m.\frac{v^2}{r} =\mu.N[/tex]
where:
[tex]\mu=[/tex] coefficient of friction
[tex]N=[/tex] normal reaction force due to weight of the car
[tex]v=[/tex] velocity of the car
[tex]1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)[/tex]
[tex]v=12.65\ m.s^{-1}[/tex] is the maximum velocity at which the vehicle can turn without skidding.
Final answer:
The maximum speed at which a car can safely make a turn on a flat road, given the mass of the car, the turn's radius of curvature, and the coefficient of friction, is approximately 25 m/s. This calculation demonstrates that the car's load does not influence its ability to negotiate the turn safely on a flat surface.
Explanation:
The student is asking how to calculate the maximum speed at which a car can safely make a turn on a flat road, given the car's mass, the turn's radius of curvature, and the coefficient of friction between the tires and the road. First, let's denote the mass of the car as m, gravitational acceleration as g, the radius of curvature as R, and the coefficient of friction as μ. To find the maximum speed v, we use the fact that the centripetal force needed to make the turn must be less than or equal to the maximum static friction force, which is μmg. This gives the condition mv²/R ≤ μmg. Solving for v, we find v = √(μgR).
By plugging in the numbers: m = 1350 kg, R = 71 m, g = 9.8 m/s², and μ = 0.23, we get v = √(0.23 * 9.8 * 71) which calculates to be approximately 25 m/s. Note, because coefficients of friction are approximate, the answer is given to only two digits.
This result is quite significant as it shows that the maximum safe speed is independent of the car's mass due to the proportional relationship between friction and normal force, which in turn is proportional to mass. This implies that how heavily loaded the car is does not affect its ability to negotiate the turn, assuming a flat surface.
You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrical wire, using all the metal, with a resistance of 1.3 Ω. How long will your wire be? What will be its diameter? The resistivity of copper is 1.7 x 10-8 Ωm. The mass density of copper is 8.96 g/cm3.
Answers
Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:
[tex]l[/tex] is the length
[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]
We divide the first equation by the second equation to get:
[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]
[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]
Using this Area, we find the diameter of the wire:
[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]
[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]
[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]
To find the length, we multiply the two equations stated initially:
[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]
[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]
At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 6 P.M.? (Round your answer to one decimal place.)
Answers
Answer:
The distance between the ships changing at 6PM is 21.29Km/h
Explanation:
Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h
Given
dx/dt= 35
dy/dt= 25
dv/dt= ???? at t= 6PM - 2PM= 4
Therefore t=4
We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction
So, we use:
[tex] D^2 = (150 - x)^2 + y^2 [/tex];
[tex] D^2 = (150 - 140)^2 + y^2 [/tex]
But ship B travels at t=4, at 4.25 =100 in the y-direction
so, let's use the equation:
[tex] D^2 = 10^2 + 100^2 [/tex]
[tex] = D= sqrt*(10 + 100) [/tex]
Lets use 2DD' = 2xx' + 2yy'
Differentiating with respect to t we have:
D•d(D)/dt = -(10)•dx/dt + 100•dy/dt
=100.5 d(D)/dt = (-10)•35 + (100)•25
When t=4, we have x=(140-150) =10 and y=100
[tex]= D = sqrt*(10^2 + 100^2) [/tex]
=100.5
= 100.5 dD/dt = 10.35 +100.25
= dD/dt = 21.29km/h
The distance between Ship A and Ship B is changing at approximately 21.4 km/h at 6 P.M.
Here, we use related rates. Let's define the positions of the ships -
Ship A's position at 6 P.M. :
Since Ship A sails east at 35 km/h for 4 hours, it would have travelled 140 km east. Initial position is 150 km west of Ship B, so at 6 P.M., Ship A will be 10 km west of Ship B.Ship B's position at 6 P.M. :
Since Ship B sails north at 25 km/h for 4 hours, it will have travelled 100 km north.Let’s denote the distance between the two ships at time t as D, x as the east-west distance (positive east) between Ship A and Ship B, and y as the north-south distance (positive north) from the initial position of Ship B.
Given:
x = -10 km, y = 100 km at 6 P.M.[tex]\frac{dx}{dt} = 35 \, \text{km/h} \quad (\text{Rate of change in east-west direction for Ship A}) \\[/tex][tex]\frac{dy}{dt} = 25 \, \text{km/h} \quad (\text{Rate of change in north-south direction for Ship B}) \\[/tex]We use the Pythagorean theorem to relate x, y, and D:
[tex]D^2 = x^2 + y^2 \\[/tex]Differentiating both sides with respect to t:
[tex]2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\[/tex]Simplifying and solving for [tex]\frac{dD}{dt}[/tex] gives:
[tex]\frac{dD}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{D} \\[/tex]Plugging in the values:
[tex]D = \sqrt{x^2 + y^2} = \sqrt{(-10)^2 + 100^2} = \sqrt{100 + 10000} = \sqrt{10100} \approx 100.5 \, \text{km} \\[/tex][tex]\frac{dD}{dt} = \frac{(-10)(35) + (100)(25)}{100.5} \approx \frac{2500 - 350}{100.5} \approx \frac{2150}{100.5} \approx 21.4 \, \text{km/h}[/tex]Therefore, the distance between the ships is changing at approximately 21.4 km/h at 6 P.M.
An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 35 m.
Answers
Answer:
10044 N
Explanation:
The acceleration of the cab is calculated using the equation of motion:
[tex]v^2 = u^2+2as[/tex]
v is the final velocity = 0 m/s in this question, since it is brought to rest
u is the initial velocity = 10 m/s
a is the acceleration
s is the distance = 35 m
[tex]a = \dfrac{v^2-u^2}{2s} = \dfrac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2\times (35\text{ m})} = -1.43\text{ m/s}^2[/tex]
Since it accelerates downwards, its resultant acceleration is
[tex]a_R = g + a[/tex]
g is the acceleration of gravity.
[tex]a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2[/tex]
The tension in the cable is
[tex]T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}[/tex]
Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a constant speed of 30 mi /hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi /hr for the positions represented.
Answers
Answer:
Incomplete question
Check attachment for the diagram of the problem.
Explanation:
The acceleration of the car A is given as
a=3ft/s²
Car B is rounding a curve of radius
r=440ft
Car B is moving at constant speed of Vb=30mi/hr.
Car A reach a speed of 45mi/hr
Note, 1 mile = 5280ft
And 1 hour= 3600s
Then
Va=45mi/hr=45×5280/3600
Va=66ft/s
Also,
Vb=30mi/hour=30×5280/3600
Vb=44ft/s
Now,
a. Let write the relative velocity of car B, relative to car A
Vb = Va + Vb/a
Then,
Using triangle rule, because vectors cannot be added automatically
Vb/a²= Vb²+Va²-2Va•VbCosθ
From the given graphical question the angle between Va and Vb is 60°.
Vb/a²=44²+66² - 2•44•66Cos60
Vb/a²=1936+ 4356 - 5808Cos60
Vb/a² = 3388
Vb/a = √3388
Vb/a = 58.21 ft/s
The direction is given as
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Va/SinA = Vb/SinB = (Vb/a)/SinC
66/SinA = 44/SinB = 58.21/Sin60
Then, to get B
44/SinB = 58.21/Sin60
44Sin60/58.21 = SinB
0.6546 = SinB
B=arcsin(0.6546)
B=40.89°
b. The acceleration of Car B due to Car A.
Let write the relative acceleration of car B, relative to car A.
Let Aa be acceleration of car A
Ab be the acceleration of car B.
Ab = Aa + Ab/a
Given the acceleration of car A
Aa=3ft/s²
Then to get the acceleration of car B, using the tangential acceleration formular
a = v²/r
Ab = Vb²/r
Ab = 44²/440
Ab = 4.4ft/s²
Using cosine rule again as above
Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ
Ab/a²= 3²+4.4²- 2•3•4.4•Cos30
Ab/a²= 9+19.36 - 22.863
Ab/a² = 5.497
Ab/a = √5.497
Ab/a = 2.34ft/s²
To get the direction using Sine rule again, as done above
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Aa/SinA = Ab/SinB = (Ab/a)/SinC
3/SinA = 4.4/SinB = 2.34/Sin30
Then, to get B
4.4/SinB = 2.34/Sin30
4.4Sin30/2.34 = SinB
0.9402 = SinB
B=arcsin(0.9402)
B=70.1°
Since B is obtuse, the other solution for Sine is given as
B= nπ - θ. , when n=1
B=180-70.1
B=109.92°
To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. The velocity of car B as observed by the observer in car A is approximately 29955/176 ft/sec. The acceleration of car B as observed by the observer in car A is approximately 1/23966164627200 mi^2/s^2.
Explanation:To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. Car B is rounding a curve at a constant speed, so its velocity remains constant. However, the observer in car A will perceive car B as having a different velocity and acceleration. The velocity of car B to the observer in car A will depend on the relative motion between the two cars, while the acceleration of car B to the observer in car A will depend on the change in direction of car B's motion.
Let's calculate the velocity and acceleration of car B as observed by an observer in car A:
Velocity: Since car B is rounding a curve with a radius of 440 ft and a constant speed of 30 mi/hr, we can use the formula v = rω to find the angular velocity ω. The angular velocity ω is equal to the speed divided by the radius, so ω = (30 mi/hr) / (440 ft) = (30 mi/hr) / (5280 ft/mi) / (440 ft) = 1/1760 rad/sec. The observer in car A will perceive car B's velocity as the vector sum of its actual velocity in the curve (tangent to the curve) and the observer's velocity in the direction of the curve (opposite to the centripetal force). Since car A has reached a speed of 45 mi/hr, its velocity can be converted to ft/sec as (45 mi/hr) / (5280 ft/mi) = 15/176 ft/sec. Therefore, the velocity of car B as observed by the observer in car A will be (30 mi/hr) + (15/176 ft/sec) = (660/22 ft/sec) + (15/176 ft/sec) = (660/22 + 15/176) ft/sec = (29955/176) ft/sec.
Acceleration: Since car B is rounding a curve at a constant speed, its acceleration is directed towards the center of the curve and has a magnitude of v^2 / r, where v is the velocity and r is the radius. Substituting the values, we get the acceleration as (30 mi/hr)^2 / (440 ft) = ((30 mi/hr)^2) / ((5280 ft/mi) / (440 ft)) = (900 mi^2/hr^2) / (5280 ft/mi) * (440 ft) = (900 mi^2 * ft^2) / (5280 hr^2) * (440) ft = (900 * 5280 * 440) ft^2 / hr^2 = (2119680000/5280) ft^2 / hr^2 = (400800 ft^2/hr^2) = (400800 ft^2/hr^2) * (1/3600 hr^2/s^2) * (1 mi^2 / (5280 ft)^2) = (400800 / 3600) * (1/5280)^2 mi^2/s^2 = (111/9900) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (1/266611200) mi^2/s^2 = (1/266611200) * (5280 ft/mi)^2 = (1/266611200) * 5280^2 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/19) ft^2/s^2 = (1/19) * (1/5280)^2 mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/26268580800) mi^2/s^2 = (1/26268580800) * (5280 ft/mi)^2 = (1/26268580800) * 5280^2 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/237896) ft^2/s^2 = (1/237896) * (1/5280)^2 mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/23966164627200) mi^2/s^2.
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A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\
Answers
Complete Question:
A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. b) How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\
b) what is the speed?
Answer:
a) the kinetic energy gained = 6.42 * 10⁻¹⁶ J
b) the speed of the particle, v = 619328.3 m/s
Explanation:
q = 1.602 *10⁻¹⁹C
V = 4.01 kV = 4.01 * 10³ V
Work done by the deuteron = qV
Work done by the deuteron = 1.602 * 10⁻¹⁹ * 4.01 *10³
Work done = 6.42 * 10⁻¹⁶ J
Kinetic Energy gained = work done
Kinetic Energy gained by the deuteron = 6.42 * 10⁻¹⁶ J
B) The formula for Kinetic Energy is given by:
KE = 1/2 Mv²
Let the mass of the proton be m₁ = 1.67262 × 10⁻²⁷kg
Let the mass of the neutron be m₂ = 1.67493 × 10−27 kg
M = m₁ + m₂
KE = 1/2 ( m₁ + m₂)v²
Let v = speed of the deuteron
From part (a)
KE = 6.42 * 10⁻¹⁶ J
1/2 ( m₁ + m₂)v²= 6.42 * 10⁻¹⁶
0.5 * (1.67262 + 1.6749) *10⁻²⁷ * v² = 6.42 * 10⁻¹⁶
v = 619328.3 m/s