Answer:
(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
Explanation:
The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"
(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
Where:
[tex]U_{g,o}[/tex], [tex]U_{g,f}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]m[/tex], [tex]M[/tex] - Masses of the rocket and planet Earth, measured in kilograms.
[tex]G[/tex] - Universal gravitation constant, measured in newton-square meters per square kilogram.
[tex]r_{o}[/tex], [tex]r_{f}[/tex] - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.
The initial distance and rocket mass are converted to meters and kilograms, respectively:
[tex]r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)[/tex]
[tex]r_{o} = 6,437,360\,m[/tex]
[tex]m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)[/tex]
[tex]m = 7000\,kg[/tex]
Given that [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]r_{f} \rightarrow +\infty[/tex], the work equation is reduced to this form:
[tex]U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}[/tex]
[tex]U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}[/tex]
[tex]U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J[/tex]
4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.
(b) The needed change in gravitational potential energy is:
[tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex]
The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:
[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]
[tex]\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}[/tex]
[tex]\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}[/tex]
[tex]r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}[/tex]
If [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex], then:
[tex]r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}[/tex]
[tex]r_{f} \approx 12,874,502.49\,m[/tex]
The final distance with respect to the center of the Earth in miles is:
[tex]r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)[/tex]
[tex]r_{f} = 7999.865\,mi[/tex]
The distance travelled by the rocket is: ([tex]r_{f} = 7999.865\,mi[/tex], [tex]r_{o} = 4000\,mi[/tex])
[tex]\Delta r = r_{f}-r_{o}[/tex]
[tex]\Delta r = 7999.865\,mi - 4000\,mi[/tex]
[tex]\Delta r = 3999.865\,mi[/tex]
The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
Where do most metamorphic processes take place?
Answer:
Most metamorphic processes take place deep underground, inside the earth's crust.
Explanation:
During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)
Equal but opposite charges Q are placed on the square plates of an air-filled parallel-plate capacitor. The plates are then pulled apart to twice their original separation, which is small compared to the dimensions of the plates. Which of the following statements about this capacitor are true?
A) The energy stored in the capacitor has doubled.
B) The energy density in the capacitor has increased.
C) The electric field between the plates has increased.
D) The potential difference across the plates has doubled.
E) The capacitance has doubled.
Answer:
A& D
Explanation:
See attached file
A rectangular loop of wire carries current I in the clockwise direction. The loop is in a uniform magnetic field B that is parallel to the plane of the loop, in the direction toward the bottom of the page. The length of the rectangle is b and the width is a. What is the net force on the loop by the magnetic field
Answer:
Explanation:
Area of the loop = a b
current = I
magnetic moment of the loop M = area x current
= ab I
Torque on the loop = MB sinθ
here θ = 90
Torque = MB
= abIB
In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose screen is 65 cm away. What refractive power will these glasses require?
Answer:
Explanation:
Far point = 17 cm . That means he can not see beyond this distance .
He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula
1 / v - 1 / u = 1 / f
1 / - 17 - 1 / - 65 = 1 / f
= 1 / 65 - 1 / 17
= - .0434 = 1 / f
power = - 100 / f
= - 100 x .0434
= - 4.34 D .
Refractive power is the measure of degree of convergence by a lens. The required refractive power for the given glasses is -4. 34 D.
Using lens formula
[tex]\bold { \dfrac 1 v - \dfrac1 u = \dfrac {1}f}[/tex]
Where,
f- focal point
v - distance of the image
u - distance of the object
So,
[tex]\bold { \dfrac 1 {-17} - \dfrac1 {-65} = \dfrac {1}f}\\\\\bold { 0.434 = \dfrac {1}f}\\[/tex]
Since, [tex]\bold {power = \dfrac {- 100 }f}[/tex]
So,
[tex]\bold { power = - 100 \times 0.0434}}\\\\\bold { power = - 4.34\ D}[/tex]
Therefore, the required refractive power for the given glasses is -4. 34 D.
To know more about refractive power,
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A 43.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.05R.
Required:
a. What is the speed of the car at the top of the vertical loop?
b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?
Answer:
A.) 909 cm/s
B.) 33075 N
Explanation:
A.) Given that the
Mass M = 43 g
Height h = 4.05 R
Radius r = R
At the top of the loop, the maximum potential energy P.E = mgh
Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2
P.E = 43 × 9610.517 × 4.05R
P.E = 1673671.536R J
According to conservative of energy
The maximum P.E = maximum K.E
But K.E = 1/2mv^2
1673671.536R = 1/2mv^2
Substitutes for mass m into the formula
1673671.536R = 1/2× 4.05R × v^2
The R will cancel out
Cross multiply
4.05 v^2 = 3347343.072
V^2 = 3347343.072 / 4.05
V^2 = 826504.4622
V = sqrt( 826504.4622)
V = 909 cm/s
B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,
MV^2/R = N + W
Make N the subject of formula
N = mv^2/ R - W
Where W = mg
Substitute all the parameters into the formula
N = (4.05R × 909^2) /R - 4.05R × 9610.517
N = 3346438.05 - 38922.59
N = 3307515 N
If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mg of Hg (= 16 kP = 1.6 × 104 N/m2). Give your answer in mg of Hg.
Answer:
The correct answer is 88.84 mmHg.
Explanation:
The pressure differential between the brain and the heart while standing up will be 120 - rho × g (gravity) × h, here h is the distance from the brain to the heart. The h is 40 cm or 0.4 m.
rho×g×h = 1060 kg/m³×9.8 m/s²×0.4m
= 4155 Pa
Now converting Pa to mmHg we get:
4155 Pa × 760 mmHg / 1.01325 × 10⁵ Pa
= 31.16 mmHg
Thus, the pressure in the brain now is 120 - 31.16
= 88.84 mmHg (hypotension)
A charged particle moves through a region of space at constant speed, without deflecting. From this, one can conclude that in this region: Select the correct answer
a) There must be no magnetic field in the region There could be electric and magnetic fields, oriented perpendicular to each other.
b) There could be electric and magnetic fields, oriented in opposite directions There must be no magnetic field and no electric field in the region.
c) There could be electric and magnetic fields, oriented in the same direction Your Ans
Answer:
There could be electric and magnetic fields, oriented in opposite directions
Explanation:
Lorentz force, is the force that may be exerted on a body of a specified magnitude of charge q, moving with a velocity v, in a magnetic field B and in an electric field of intensity E. This Lorentz force is given by; F= qE+qvBsin ϕ
However, if the motion of the particle is opposite to the magnetic field such a that ϕ = 0, then there is no net magnetic force on the charge and it moves freely, with a constant velocity and in a straight line. Hence, there is no magnetic field in the region.
The charge moves with constant speed due to same direction of magnetic and electric field.
There could be electric and magnetic fields that is oriented in the same direction or the other reason is that there is no magnetic field and electric field in that region where the charge moves. If the electric and magnetic field are present at the same direction then it means that it applies no force on the charge.
This is due to more distance from the charge as well as the charge travels away from the field occupies by the magnetic and electric field so we can conclude that the charge moves with constant speed due to same direction of magnetic and electric field.
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A microwave oven uses 1,200 watts. It runs a total of 6 hours during the corse of a week. How much energy does the microwave oven use during the week?? a 4.8 kilowatt-hours. b 5.0 kilowatt-hours. c 7.2 kilowatt-hours. d 6.5 kilowatt-hours.
Answer:
7.2 kilowatts
Explanation:
Energy= power x time
where:
power = 1,200watts to kilowatts is 1.2kw
time = 6hours
therefore,
energy = 1.2kw x 6hrs
=7.2 kilowatts
Science activity
Imagine that some settlers have left Earth and gone to the Moon, taking
their recipe books with them. The first cake they baked was a disaster. It had
far too little moisture and was about six times the size they had expected.
the cake recipe was:
1.25 N butter
1.50 N sugar
4 eggs
1.50 N flour
20 ml milk
ANALYTICAL THINKING
Q. Why was the cake so big? Why was it se
dry?
Answer:
Answer in explanation
Explanation:
The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:
m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg
But, on moon:
m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg
Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.
Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.
explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage
Answer:
polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.
the cup helps it become more insulated
The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of its full equilibrium charge
Answer:
2.7s
Explanation:
The solution of time required is shown below:-
In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time
Now, the
63% that is equal to 0.63 which is full equilibrium charge
Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s
So, the correct answer is 2.7s
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
Learn more:
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What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration of the box?
Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
Which one of the conditions can cause a particle to move with uniform circular motion in a uniform magnetic field
Given that,
A particle to move with uniform circular motion in a uniform magnetic field.
Suppose, The conditions are,
(I). The charged particle has to be positive and it should be moving in a direction opposite to a uniform magnetic field.
(II). The charged particle should be moving parallel to the magnetic force and perpendicular to the magnetic field.
(III). The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
We know that,
An particle to move with uniform circular motion.
Here, electric force is perpendicular to velocity of particle.
The electric field is defined as,
[tex]F_{c}=\dfrac{mv^2}{r}[/tex].....(I)
Suppose, there is magnetic field, if a charge moving with velocity and the magnetic field exerts a field.
The magnetic force is defined as,
[tex]F_{m}=qvB[/tex].....(II)
We need to find the magnetic field
Using equation (I) and (II)
[tex]F_{c}=F_{m}[/tex]
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
Hence, The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
(III) is correct option.
When you replace helium in a balloon with less-dense hydrogen, does the buoyant force on the balloon change if the balloon remains the same size?
Answer:
No change
Explanation:
First, we hare to understand what we mean by buoyant force.
Archimedes Principle states that
"the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object."
Hot air balloons rise into the air as a result of the density of the air inside the balloon is less dense i.e warmer air than the air outside the balloon i.e cooler air. This is basically how the balloons work. Now applying this to our question
Replacing Helium with less dense Hydrogen will make no difference to the buoyant force because the volume of the balloon did not change. The buoyant force depends on the weight of the displaced air, and not on the force causing the displacement.
A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J
Answer:
Explanation:
v₁² = v₁ . v₁
= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )
= 12.25 + 8.41
= 20.66 m /s
a ) kinetic energy = 1/2 m v₁²
= 1/2 x .7 x 20.66
= 7.23 J
b )
changed velocity v₂ = v₂.v₂
= (6i - 5 j ) . (6i - 5 j )
= 36 + 25
= 61 m /s
kinetic energy = 1/2 m v₂²
= 1/2 x .7 x 61
= 21.35 J
Work done = change in energy
= 21.35 - 7.23
= 14.12 J .
A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true?
a. the mass is initially released from rest from a point 1 meter below the equilibrium position
b. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 11 m/s
Answer:
Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.
a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
Explanation:
Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)
[tex]x^\prime[/tex] would thus represent the velocity of this mass.[tex]x^{\prime\prime}[/tex] would represent the acceleration of this mass.Constructing the ODEConstruct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.
The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:
[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].
The question described another equation for the net force on this mass. This equation is the sum of two parts:
The restoring force of the spring: [tex]F(\text{spring}) = -k\, x[/tex], where [tex]k[/tex] denotes the constant of this spring.The damping force: [tex]F(\text{damping}) = - 11\,x^\prime[/tex] according to the question. Note the negative sign in this expression- the damping force should always oppose the direction of motion.Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:
[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].
Combine the two equations for the net force on this mass to obtain:
[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].
From the question:
Size of this mass: [tex]m = 1\; \rm kg[/tex].Spring constant: [tex]k = 18\; \rm N \cdot m^{-1}[/tex].Hence, the equation will become:
[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].
Rearrange to obtain:
[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].
Finding the general solution to this ODE[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:
[tex]m^2 + 11\, m + 18 = 0[/tex].
The two roots are:
[tex]m_1 = -2[/tex], and[tex]m_2 = -9[/tex].Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:
[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].
For this particular ODE, that general solution would be:
[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].
Finding the particular solutions to this ODENote, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time
The position at time [tex]t = 0[/tex] would be [tex]x(0) = c_1 + c_2[/tex].The velocity at time [tex]t = 0[/tex] would be [tex]x^\prime(0) = -2\, c_1 - 9\, c_2[/tex].For section [tex]\rm a.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm a.[/tex] will be:
[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
Similarly, for section [tex]\rm b.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm b.[/tex] will be:
[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
There are fiber optic telephone cables connecting North America and Europe, lying on the bottom of the Atlantic ocean. The wire is 4,500 km long how long and has an index of refraction of 1.5. How long will it take for the signal to cross the ocean? Give your answer in milliseconds.
Answer:
The time taken is [tex]t = 0.0225 \ s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]
The refractive index is [tex]n_f = 1.5[/tex]
The velocity of the signal is mathematically represented as
[tex]v = \frac{c}{n_f }[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]
[tex]v = 2.0*10^{8} \ m/s[/tex]
The time taken is mathematically evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]
[tex]t = 0.0225 \ s[/tex]
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
When a ray of light traveling in air hits a tilted plane parallel slab (of glass, say), it emerges parallel to the original ray but shifted transversely. Carefully draw out the situation and use Snell’s law to derive the amount of the transverse shift, x, as a function of the tilt angle of the slab, θ, its thickness, d, and its index of refraction, n. Find the exact expression with no approximations. We recommend you do this out all in variables because it's a useful formula to have. Also, you will want this for the following questions. However, since the auto-grader has difficulty with these formulas, use n=1.5, d=1.0 cm, and θ = 45° and enter a numerical answer. Give your answer in cm to two significant figures.
Answer:
x = 0.4654 cm
Explanation:
In this exercise we use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
apply this formula to the first surface, where n₁ is the index of refraction of air (n₁ = 1) and n₂ is the index of refraction of glass (n₂ = n)
θ₂ = sin⁻¹ (sin θ₁ / n) (1)
having this angle we use trigonometry to find the value of the point where it comes out when we reach the other side
refracted ray
tan θ₂ = x₂ / d
x₂ = d tan θ₂
this value is the distance displaced by the refracted ray
now let's find the distance at which the incident beam should exit
tan θ₁ = x₁ / d
x₁ = d tan θ₁
the displacement of the ray is the difference between these two distances, we will call it x
x = x₁ - x₂
x = d tan θ₁ - d tan θ₂
x = d (tan θ₁ - tan θ₂) (2)
the easiest way to do the calculations is to find tea2 from the binding 1 and then perform the calculation with equation 2
calculate
θ₂ = sin⁻¹ (sin 45 /1.5)
θ₂ = 28.13º
x = 1.0 (tan 45 - tan 28.13)
x = 0.4654 cm
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.0050 m) is attached to the end of the hose in units of m/s.
Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
v₂ = 306.12 m/s
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through
Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
calculate the upthrust aciting on a body if its
true weight is 550 N and apparent weight
lis 490 N
Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
Learn more about force here https://brainly.com/question/25239010
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Find an article online or application in your daily life involving rotating objects and physics.
Answer:
the planet Earth is a good example