Answer: A BIRD BRAINLIEST PLEASE
Explanation: If a plane was traveling at the same velocity as a bird, which would have the most kinetic energy (assuming the plane has more mass)? ... So if a 747 weighs 750,000 times as much as a bird, at the same velocity it will have 750,000 times the kinetic energy.
The two objects have the same speed, thus the plane will have more kinetic energy compared to the bird.
Kinetic energy is the energy acquired by a body during its motion. This kinetic energy depends on the speed and mass of the object.
The formula for estimating kinetic energy is given as;
K.E = ¹/₂mv²
where;
m is mass of the objectv is the speed of the objectThe mass of the plane should be greater than the mass of the bird. Since the two objects have the same speed, we can conclude that the plane will have more kinetic energy compared to the bird.
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A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. When it returns to the surface it is determined that at the bottom of the lake the water rose to within 0.133 meters of the closed end. The lake is 100 meters deep, the air temperature at the surface is 27 "C, atmospheric pressure is 1.01x10s N/m2, and the density of water is 998 kg/m3. a) What is the total pressure at the bottom of the lake
Complete Question
The complete question is shown on the first uploaded image
Answer:
The total pressure is [tex]P_T = 10.79*10^{5} N/m^2[/tex]
The temperature at the bottom is [tex]T_b = 284.2 \ K[/tex]
Explanation:
From the question we are told that
The length of the glass tube is [tex]L = 1.50 \ m[/tex]
The length of water rise at the bottom of the lake [tex]d = 1.33 \ m[/tex]
The depth of the lake is [tex]h = 100 \ m[/tex]
The air temperature is [tex]T_a = 27 ^oC = 27 +273 = 300 \ K[/tex]
The atmospheric pressure is [tex]P_a = 1.01 *10^{5} N/m[/tex]
The density of water is [tex]\rho = 998 \ kg/m^3[/tex]
The total pressure at the bottom of the lake is mathematically represented as
[tex]P_T = P_a + \rho g h[/tex]
substituting values
[tex]P_T = 1.01*10^{5} + 998 * 9.8 * 100[/tex]
[tex]P_T = 10.79*10^{5} N/m^2[/tex]
According to ideal gas law
At the surface the glass tube not covered by water at surface
[tex]P_a V_a = nRT_a[/tex]
Where is the volume of
[tex]P_a *A * L = nRT_a[/tex]
At the bottom of the lake
[tex]P_T V_b = nRT_b[/tex]
Where [tex]V_b[/tex] is the volume of the glass tube not covered by water at bottom
and [tex]T_b[/tex] i the temperature at the bottom
So the ratio between the temperature at the surface to the temperature at the bottom is mathematically represented as
[tex]\frac{T_b}{T_a} = \frac{d * P_T}{P_a * h}[/tex]
substituting values
[tex]\frac{T_b}{27} = \frac{0.133 * 10.79 *10^5}{1.01 *10^{5} * 1.5}[/tex]
=> [tex]T_b = 284.2 \ K[/tex]
A vehicle travels 2345 m in 315 s toward the evening sun. What is it's speed
Answer:
7.44 m/s
Explanation:
Units
The first thing you must do is get the units.
The distance is in meters.
The time is in seconds.
Therefore the speed is going to be in meters/second.
Givens
d = 2345 meters
t = 315 seconds
Solution
s = d/t
s = 2345/315
s = 7.44 meters/second
how many grams of nitrogen gas are needed to produce 34 g of ammonia
Answer:
28 grams
Explanation:
The equation for the reaction is
3H(2) + N(2) -> 2NH(3)
Then we have.
The molar mass, M of ammonia is 17 g/mol.
34 grams of ammonia, NH3 then would be
34 g / 17 g/mol
= 2 moles
2 moles of ammonia will be obtained from
(2 * 1) / 2
= 1 mole of nitrogen
The molar masses of nitrogen is 28 g/mol
2 moles of nitrogen corresponds to 1 * 28 = 28 grams.
help asap. which of the following use the most energy?
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:
garage door openers: 40.0 MHz
standard cordless phones: 40.0 to 50.0 MHz
baby monitors: 49.0 MHz
FM radio stations: 88.0 to 108 MHz
cell phones: 800 to 900 MHz
Global Positioning System: 1227 to 1575 MHz
microwave ovens: 2450 MHz
wireless internet technology: 2.4 to 2.6 GHz
Which of the following statements correctly describe the various applications listed above? Check all that apply.
a.) All these technologies use radio waves, including low-frequency microwaves.
b.) All these technologies use radio waves, including high-frequency microwaves.
c.) All these technologies use a combination of infrared waves and high-frequency microwaves.
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.
The correct statements are: (b) All these technologies use radio waves, including high-frequency microwaves. (d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
Explanation:The correct statements that describe the various applications listed above are:
b.) All these technologies use radio waves, including high-frequency microwaves.d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.Statement a.) is incorrect because not all technologies listed use low-frequency microwaves. Statement c.) is incorrect because not all technologies listed use a combination of infrared waves and high-frequency microwaves. Statement e.) is incorrect because wireless Internet devices do not have the shortest wavelength among the technologies listed. Statement f.) is incorrect because the wavelengths of the technologies listed vary. Statement g.) is incorrect because the wavelengths of the technologies listed also vary.
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The correct answers are option (a) and option (d). All these technologies use radio waves, including low-frequency microwaves and Microwave ovens emit in the same frequency band as some wireless Internet devices.
The electromagnetic (EM) spectrum includes a wide range of frequencies, which are used in various modern technologies. Here are analyses correlating with the provided frequency ranges:
All these technologies use radio waves, including low-frequency microwaves: This statement is correct because the frequency ranges provided fall within the radio wave section of the EM spectrum (300 kHz to 300 GHz).All these technologies use radio waves, including high-frequency microwaves: This statement is partially correct because they indeed use radio waves, but not all fall under high-frequency microwaves.All these technologies use a combination of infrared waves and high-frequency microwaves: This statement is incorrect as none of the mentioned applications utilize infrared waves.Microwave ovens emit in the same frequency band as some wireless Internet devices: This is correct. Microwave ovens operate at 2450 MHz, which overlaps with wireless internet technologies operating at 2.4 GHz (or 2400 MHz).The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above: This is incorrect. Different technologies listed operate within varying ranges, some of which have shorter wavelengths.All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m: This is incorrect because not all of the frequencies provided correspond to this specific wavelength range.All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km: This is also incorrect since the provided frequencies exceed these wavelength ranges.Tish scarf is 17.75 inches long and knits 2 3/8 per minute. Emma scarf is 4 inches knits 3 3/4 inches per minute. After how many minutes will Emma’s scarf be longer than tish scarf
Answer:
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
Explanation:
Given;
The length of tish scarf after t minutes can be written as;
L1 = 17.75 + 2 3/8 ×t
L1 = 17.75 + 2.375t
The length of emma scarf after t minutes can be written as;
L2 = 4 + 3 3/4 × t
L2 = 4 + 3.75t
For emma scarf to be longer than tish;
L2 >/= L1
At L2 = L1
4+3.75t = 17.75 + 2.375t
Solving for t
3.75t - 2.375t = 17.75 - 4
1.375t = 13.75
t = 13.75/1.375
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
Katie rubs a balloon against her hair. Electrons from her hair travel to the balloon, giving the balloon a negative charge and her hair a positive charge.
When the negatively charged balloon is brought near the strands of Katie's hair, they move to get closer to the balloon, without the balloon actually touching them. This shows that
A. electric attraction is a force that can only act on contact.
B. Katie's hair would move in this way with or without the balloon.
C. electric attraction is a force that can act at a distance.
D. particles in the air must be pulling Katie's hair toward the balloon.
Answer:
c. electric attraction is a force that can act at a distance.
Explanation:
stuisland
Final answer:
A negatively charged balloon attracting positively charged hair strands without contact illustrates that c. electric attraction can act at a distance.
Explanation:
When Katie rubs the balloon against her hair, electrons move from her hair to the balloon, resulting in the balloon having a negative charge and her hair having a positive charge. If the negatively charged balloon is brought near Katie's hair and the hair strands move toward it without direct contact, this demonstrates electric attraction as a force that can act at a distance.
Therefore, electric attraction is a force that can act at a distance. This example, similar to the effect observed when someone touches a Van de Graaff generator, shows charge separation and induction. It validates the scientific concept that electric forces can operate between charged objects even when they are not in physical contact.
In an ultrahigh vacuum system, the pressure is measured to be 8.4 × 10−11 torr (where 1 torr = 133 Pa). The gas molecules have a molecular diameter of 2.2 × 10−10 m and the temperature is 310 K. Avogadro's number is 6.02214×1023 1/mol. Find the number of molecules in a volume of 0.87 m3 . Answer in units of molecules.
Answer:
The number of molecules in the volume is [tex]N_v = 2.27109* 10^{12}[/tex] molecules
Explanation:
From the question we are told that
The pressure of the ultrahigh vacuum is [tex]P = 8.4*10^{-11} torr = 8.4*10^{-11} * 133 = 1.1172 *10^{-8}Pa[/tex]
The molecular diameter of the gas molecules [tex]d = 2.2*10^{-10} m[/tex]
The temperature is [tex]T = 310 \ K[/tex]
Avogadro's number is [tex]N = 6.02214 *10^{23}\ l/mol[/tex]
The volume of the gas is [tex]V = 0.87 m^3[/tex]
From the ideal gas law[[tex]PV = nRT[/tex]] that the number of mole is mathematically represented as
[tex]n = \frac{PV}{RT}[/tex]
Where R is the gas constant with a value [tex]R = 8.314\ J/mol[/tex]
Substituting values
[tex]n = \frac{1.1172 *10^{-8} * 0.87}{8.314 * 310}[/tex]
[tex]n = 3.771*10^{-12} \ mole[/tex]
The number of molecules is mathematically represented as
[tex]N_v = n * N[/tex]
Substituting values
[tex]N_v = 3.771*10^{-12} * 6.02214 *10^{23}[/tex]
[tex]N_v = 2.27109* 10^{12}[/tex] molecules
The simple pendulum above consists of a bob hanging from a light string. You wish to experimentally determine the frequency of the swinging pendulum. (a) By checking the line next to each appropriate item on the list below, select the equipment that you would need to do the experiment. ____ Meter Stick ____ Protractor ____ Additional string ____ Stopwatch ____ Photogate ____ Additional masses (b) Describe the experimental procedure that you would use. In your description, state the measurements you would make, how you would use the equipment to make them, and how you would determine the frequency from those measurements. (c) You next wish to discover which parameters of a pendulum affect its frequency. State one parameter that could be varied, describe how you would conduct the experiment, and indicate how you would analyze the data to show whether there is a dependence. (d) After swinging for a long time, the pendulum eventually comes to rest. Assume that the room is perfectly thermally insulated. How will the temperature of the room change while the pendulum comes to rest
A) the item required for the experiment is a stopwatch.
B) You are looking for time savings. To get this, divide the time by 10. The result is the period. The formula is given as
F = [tex]\frac{1}{T}[/tex] Where F is the frequency i.e. the number of cycles per second and T is the number of seconds per cycle.
C) One parameter of the pendulum that can be altered in order to affect the frequency of the pendulum is its length. A pendulum with a longer string will have a lower frequency.
The one with a shorter length will have a higher frequency
D) In an environment that is thermally insulated perfectly, it means that any heat generated within the room is trapped within it. As the pendulum comes to rest, the room will experience a slight increase in temperature due to the conversion of mechanical energy to thermal energy.
Uses of the pendulum experimentsUnderstanding the physics of pendulums helps one to get a better grasp of gravity, inertia, and centripetal force.
Pendulums are used for the construction or engineering of clocks, metronomes, sismometers, amusement park rides.
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The simple pendulum above consists of a bob hanging from a light string the experiments is :
A. Stopwatch
B. You are looking for time savings. To get this, divide the time by 10.
C. The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length.
D. In an environment that's thermally protects superbly
"Simple Pendulum"Answer A:
The item that is required for the experiment is a stopwatch.Answer B:
The experimental procedure that you would use is :
You are looking for time savings. To get this, divide the time by 10. The result is the period. Formula is given as :F = 1/10Where :
F is the frequency and T is the number of seconds per cycle.Answer C:
The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length. A pendulum with a longer string will have a lower frequency. The one with a shorter length will have a better frequency.Answer D:
In an environment that's thermally protects superbly, it implies that any warm created inside the room is caught inside it. As the pendulum comes to rest, the room will encounter a slight increment in temperature due to the change of mechanical vitality to warm energy.Uses of the pendulum experiments :
Understanding the material science of pendulums helps one to induce distant better a much better, a higher, a stronger ,an improved an improved get a handle on of gravity, inactivity, and centripetal force. Pendulums are utilized for the development or building of clocks, metronomes, sismometers, entertainment stop rides.Learn more about pendulum experiments :
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This is theDopplereffect. Sup-pose that, at a particular moment, you are in a train traveling at 34 m/s and acceleratingat 1.2m/s2. A train is approaching you from the opposite direction on the other trackat 40 m/s, accelerating at 1.4m/s2, and sounds its whistle, which has frequency of 460Hz. At that instant, what is the perceived frequency that you hear and how fast is itchanging
Complete Question
If a sound with frequency fs is produced by a source traveling along a line with speed vs and an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is
f_o = [(c+v_o)/(c-v_s)] f_s
where c is the speed of sound, about 332 m/s. (This is the Doppler effect). Suppose that, at a particular moment, you are in a train traveling at 34 m/s and accelerating at 1.2 m/s^2. A train is approaching you from the opposite direction on the other track at 40 m/s, accelerating at 1.4 m/s^2, and sounds its whistle, which has a frequency of 460Hz. At that instant, what is the perceived frequency that you hear and how fast is it changing?
Answer:
The frequency the person hears is [tex]f_o = 557 Hz[/tex]
The speed at which it is changing is [tex]\frac{df_o}{dt} = 4.655 Hz/s[/tex]
Explanation:
Form the question we are told that
The frequency of the sound produced by source is [tex]f_s[/tex]
The speed of the source is [tex]v_s[/tex]
The speed of the observer
The frequency of sound heard by observer [tex]f_o =[ \frac{c + v_o }{c - v_s} ] * f_s[/tex]
The speed of sound is c with value [tex]c = 332 m/s[/tex]
Looking the question we can deduce that the person in the first train is the observer so the
[tex]v_o = 34 m/s[/tex]
and the acceleration is [tex]\frac{dv_o}{dt} = 1.2 m/s^2[/tex]
The train the travelling in the opposite direction the blew the whistle
is the source
So [tex]v_s = 40 m/s[/tex]
and [tex]f_s = 460 Hz[/tex]
and the acceleration is [tex]\frac{dv_s}{dt} = 1.4 m/s^2[/tex]
We are told that
[tex]f_o =[ \frac{c + v_o }{c - v_s} ] * f_s[/tex]
Substituting values we have that
[tex]f_o =[ \frac{332 + 34 }{332 - v40} ] * 460[/tex]
[tex]f_o = 557 Hz[/tex]
Differentiating [tex]f_o[/tex] using chain rule we have that
[tex]\frac{d f_o}{dt} = \frac{df_o}{dt } * \frac{dv_o}{dt} + \frac{d f_o}{dv_s} * \frac{dv_s}{dt}[/tex]
Now
[tex]\frac{df_o}{dt } = \frac{f_s}{c- v_s}[/tex]
[tex]\frac{df_o}{dv_s} = \frac{c+ v_o}{c-v_s} f_s[/tex]
Substituting this into the equation
[tex]\frac{df_o}{dt} = \frac{f_s}{c-v_s} * \frac{d v_o}{dt} + \frac{c+v_o}{(c-v_s)^2} f_s * \frac{dv_s}{dt}[/tex]
Now substituting values
[tex]\frac{df_o}{dt} = \frac{460}{332 - 40} * (1.2) + \frac{332+ 34}{(332- 40)^2} 460 * 1.4[/tex]
[tex]\frac{df_o}{dt} = 4.655 Hz/s[/tex]
Vectors A and B lie in the x-y plane. Vector A has a magnitude of 17.6 and is at an angle of 120.5° counter-clockwise from the x-axis. Vector B has a magnitude of 21.7 and is 240.3° from the x-axis. Resolve A and B into components, and express in unit vector form below.
The unit vector forms of the given vectors is required.
The required vectors are [tex]A=-8.93\hat{i}+15.16\hat{j}[/tex] and [tex]B=-10.75\hat{i}-18.84\hat{j}[/tex]
VectorsMagnitude of vector A = [tex]|A|=17.6[/tex]
Angle vector A makes with positive x axis counter clockwise = [tex]\theta_1=120.5^{\circ}[/tex]
Magnitude of vector B = [tex]|B|=21.7[/tex]
Angle vector B makes with positive x axis counter clockwise = [tex]\theta_2=240.3^{\circ}[/tex]
The vectors need to be resolved in order to write in the unit vector forms.
The vectors are
[tex]A=|A|(\cos\theta_1\hat{i}+\sin\theta_1\hat{j})\\\Rightarrow A=17.6(\cos120.5\hat{i}+\sin120.5\hat{j})\\\Rightarrow A=-8.93\hat{i}+15.16\hat{j}[/tex]
[tex]B=|B|(\cos\theta_2\hat{i}+\sin\theta_2\hat{j})\\\Rightarrow B=21.7(\cos240.3\hat{i}+\sin240.3\hat{j})\\\Rightarrow B=-10.75\hat{i}-18.84\hat{j}[/tex]
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Final answer:
To resolve the vectors into components, apply trigonometric functions -- cosine for x components and sine for y components -- to their magnitudes and angles, then express them in unit vector form with i and j.
Explanation:
When resolving vectors A and B into their components in the x-y plane, the general method involves using trigonometry, specifically, the cosine and sine functions for the x and y components, respectively. Given that vector A has a magnitude of 17.6 and an angle of 120.5° from the x-axis, its components can be calculated as follows:
Ax = A * cos(θ) = 17.6 * cos(120.5°)
Ay = A * sin(θ) = 17.6 * sin(120.5°)
Similarly, vector B with a magnitude of 21.7 and an angle of 240.3° from the x-axis has components:
Bx = B * cos(θ) = 21.7 * cos(240.3°)
By = B * sin(θ) = 21.7 * sin(240.3°)
The resulting components should then be written in unit vector form by attaching the unit vectors i (for the x-axis) and j (for the y-axis) to the respective components.
A snail can crawl 160cm at an avg speed of 4cm/min. If it crawled at an avg speed of 5cm/min instead, how much sooner would it take to reach the destination?
Answer:
8 minutes sooner
Explanation:
Average speed of snail= 4cm/min
Distance to be covered = 160cm
Time taken for the journey = distance/speed
Time taken for the journey = 160/4
Time taken for the journey = 40 min
If it crawed an average speed of 5cm/min
Distance = 160 cm
Time for the journey = distance/speed
Time for the journey = 160/5
Time for the journey = 32 min
Its going to take the snail 40 min - 32 min to Kno how sooner it will taje it if the average speed is 5cm/min
40 min - 32 min = 8 min
Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?
Answer:
Explanation:
On both sides of the film , the mediums have lower refractive index.
for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is
2μt = ( 2n +1 ) λ / 2
μ is refractive index of film ,t is thickness of film λ is wavelength of light
n is order of fringe
for minimum thickness
n = 0
2μt = λ / 2
t = λ / 4μ
= 670 / 1.75 x 4
= 95.71 nm .
To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 The pupil of the eye is the circular mm in dim light. Find the angular resolution of the eye for 550 nm opening through which light enters. wavelength light at those extremes. In which light can you see more sharply, dim or bright
Correct question is;
To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?
Answer:
We'll see more sharply in dim light
Explanation:
If we consider diffraction through a circular aperture, then angular resolution is given by;
θ = 1.22λ/D
where:
θ is the angular resolution (radians) λ is the wavelength of light
D is the diameter of the lens' aperture.
Thus,
at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm
θ = (1.22 * 550)/(2 x 10^(6))
θ = 335.5 x 10^(-6) radians
Now, we need to convert this to arc seconds.
Thus;
1 arc second = 4.85 x 10^(-6) radians
So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]
= 69.18 arc seconds
at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm
θ = (1.22 * 550)/(8 x 10^(6))
θ = 83.875 x 10^(-6) radians
Now, we need to convert this to arc seconds.
Thus;
1 arc second = 4.85 x 10^(-6) radians
So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]
= 17.3 arc seconds
From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.
Consider an incident normal shock wave that reflects from the end wall of a shock tube. The air in the driven section of the shock tube (ahead of the incident wave) is at p, = 0.01 atm and TI = 300 K. The pressure ratio across the incident shock is 1050. With the use of Eq. (7.23), calculate a. The reflected shock wave velocity relative to the tube b. The pressure and temperature behind the reflected shock
Answer:
Find the given attachments for complete solution
A 969-kg satellite orbits the Earth at a constant altitude of 99-km. (a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 195 km? 469 Incorrect: Your answer is incorrect. How is the total energy of an object in circular orbit related to the potential energy? MJ (b) What is the change in the system's kinetic energy? MJ (c) What is the change in the system's potential energy?
Answer:
1.3*10^14 J
Explanation:
The energy of the satellite that orbits the earth is given by the second Newton law:
[tex]F=ma_c\\\\-G\frac{mM_s}{r^2}=m\frac{v^2}{r}\\\\v^2=\frac{GM}{r}\\\\E_T=K+U=G\frac{mM_s}{2r}-G\frac{mM}{r}=-G\frac{mM}{2r}[/tex]
where you have taken into account the centripetal acceleration of the satellite.
m: mass of the satellite
M_s: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.67*10^-11 m^3/kg s^2
r: distance to the center of the Earth = Earth radius + distance satellite-Earth surface
To find the needed energy, you first compute the energy for a constant altitude of 99km:
r = 6.371*10^6m + 99*10^3m = 6.47*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.47*10^6)}\\\\E_T=-9.88*10^{15} \ J[/tex]
Next, you calculate the energy for an altitude of 195km:
r = 6.371*10^6m + 195*10^{3}m = 6.56*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.56*10^6)}\\\\E_T=-9.75*10^{15} \ J[/tex]
Finally, the energy required to put the satellite in the new orbit is:
-9.75*10^15 J - (-9.88*10^15 J) = 1.3*10^14 J
How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?
Answer:
9.2 amperes
Explanation:
Ohm's law states that the voltage V across a conductor of resistance R is given by [tex]V = R I[/tex]
Here, voltage V is proportional to the current I.
For voltage, unit is volts (V)
For current, unit is amperes (A)
For resistance, unit is Ohms (Ω)
Put R = 12.5 and V = 115 in V=RI
[tex]115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes[/tex]
A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at time t is given by the equation v=−2 cos(3πt) with v measured in feet per second and t measured in seconds. Determine the maximum velocity of the mass and the amount of time it takes for the mass to move from its lowest position to its highest position.
Answer:
the maximum velocity of the mass v (max) = 2 ft/s
the amount of time it takes for the mass to move from its lowest position to its highest position∆t = 1/3 seconds = 0.33 seconds
Explanation:
Given the velocity equation;
v=−2 cos(3πt)
The maximum velocity would be at cos(3πt) = 1 or cos(3πt) = -1
v (max) = -2 × -1 = 2 ft/s
The time taken for the mass to move from lowest position to highest position
At Lowest position, vertical velocity equals zero.
At highest position, vertical velocity equals zero.
The time taken for the mass to move from one v = 0 to the next v = 0
Cos(π/2) = 0 and
Cos(3π/2) = 0
For the first;
Cos(3πt) = cos(π/2)
3πt1 = π/2
t1 = π/2(3π)
t1 = 1/6 second
For the second;
Cos(3πt) = cos(3π/2)
3πt2 = 3π/2
t2 = 3π/2(3π)
t2 = 1/2 second
∆t = t2 - t1 = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 seconds
∆t = 1/3 seconds
A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acceleration at the earth's surface.)
Part A
Find the maximum tension in the chain.
Tmax = N
Part B
Find the minimum tension in the chain.
Tmin = N
Part C
What is the tension at a point three-fourths of the way up from the bottom of the cha
Answer:
Tension maximum =1131.9 N
Tension minimum =868.28 N
Tension at 3/4= 1065.995 N
Explanation:
a)
Given Mass of wrecking ball M1=88.6 Kg
Mass of the chain M2=26.9 Kg
Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)
=(88.6+26.9) × (9.8 m/s²)
=115.5 × 9.8 m/s²
Tension maximum =1131.9 N
b)
Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²
=88.6 × 9.8 m/s²
Tension minimum =868.28 N
c)
Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9
= (3/4 × 26.9)+88.9) × 9.8 m/s²
= (20.175+88.6) × 9.8 m/s²
=(108.775) × 9.8 m/s²
=1065.995 N
Final answer:
The maximum tension in the chain is 1131.9 N, occurring at the top, while the minimum tension is 263.62 N at the bottom. The tension at a point three-fourths the way up from the bottom is 935.465 N.
Explanation:
To find the maximum and minimum tension in the chain, we need to consider the system's configuration, and the force due to gravity. The maximum tension occurs at the top of the chain, where it supports the entire weight of the wrecking ball and the chain. The minimum tension occurs at the bottom of the chain, where it only needs to support the chain's weight. To find the tension at a point three-fourths of the way up from the bottom, we need to consider the weight of the portion of the chain below that point and the wrecking ball's weight.
Maximum tension (Tmax) is the sum of the weight of the wrecking ball and the entire chain:
Tmax = (mass of ball + mass of chain) × gravitational acceleration
Tmax = (88.6 kg + 26.9 kg) × 9.80 m/s²
Tmax = 115.5 kg × 9.80 m/s²
Tmax = 1131.9 N
Minimum tension (Tmin) is just the weight of the chain:
Tmin = mass of chain × gravitational acceleration
Tmin = 26.9 kg × 9.80 m/s²
Tmin = 263.62 N
Tension at three-fourths the way up:
We calculate the weight of the top one-fourth of the chain plus the wrecking ball:
Tension at three-fourths the way up = (mass of one-fourth of the chain + mass of ball) × gravitational acceleration
Tension at three-fourths = ((26.9 kg / 4) + 88.6 kg) × 9.80 m/s²
Tension at three-fourths = (6.725 kg + 88.6 kg) × 9.80 m/s²
Tension at three-fourths = 935.465 N
An object is moving in the absence of a net force. Which of the following best describes the object’s motion? A. The object will slow down at a constant rate until coming to rest B. The object will stop moving and remain at rest until acted on by a net force C. The object will continue to move at a constant speed but in a circular path D. The object will continue to move with a constant velocity
Answer:
D. The object will continue to move with a constant velocity
Explanation:
According to Newton's first law also known as law of inertia, states that an object at rest will remain at rest or, if in motion, will remain in motion at constant velocity unless acted on by a net external force.
Therefore, An object moving in the absence of a net force will continue to move at a constant velocity
In the absence of a net force, an object will continue to move with a constant velocity.
Explanation:The correct answer is D. The object will continue to move with a constant velocity. In the absence of a net force, an object will continue to move at a constant velocity. This means that the object will continue to move in a straight line at the same speed without slowing down or changing direction.
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Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjacent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a = 16.3 cm a=16.3 cm from the wall. The reflected beam strikes the wall at a height b = 32.5 cm b=32.5 cm above the surface of the mirror. Find the angle of incidence θ i θi at which the laser beam strikes the mirror.
Answer:
26.64°
Explanation:
Given:
a = 16.3 cm
b = 32.5 cm
Angle when laser beam reflects off the mirror and strike the wall =
θ [tex] = tan^-^1(\frac{b}{a}) [/tex]
[tex] = tan^-^1(\frac{32.5}{16.3}) [/tex]
= 63. 36°
For angle of reflection, we have:
θr = 90° - 63.36°
θr = 26.64°
Since angle of incidence, θi is equal to angle of reflection θr, the angle of incidence θi at which the laser beam strikes the mirror is =
θi = θr = 26.64°
Three balls with the same radius 21 cm are in water. Ball 1 floats, with
half of it exposed above the water level. Ball 2, with a density 893
kg/m3 is held below the surface by a cord anchored to the bottom of the
container, so that it is fully submerged. Ball 3, of density 1320 kg/m3, is
suspended from a rope so that it is fully submerged. Assume the
density of water is 1000 kg/m3 in this problem.
A. Which is true for Ball 1?
B. What is the tension on the rope holding the second ball, in newtons?
C. What is the tension on the rope holding the third ball in N?
Ball 1 floats with half exposed above water level. Tension on rope holding Ball 2 is calculated using weight and buoyant force. Tension on rope holding Ball 3 is equal to buoyant force.
Explanation:A. Ball 1 is floating with half of it exposed above the water level.
This means that the buoyant force on the ball is equal to the weight of the ball.
Since the buoyant force is greater than the weight of the ball, the ball floats.
B. The tension on the rope holding Ball 2 can be found using the equation:
Tension = Weight - Buoyant force.
The weight of the ball is calculated by multiplying its volume by its density and acceleration due to gravity.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
C. The tension on the rope holding Ball 3 is the same as the buoyant force acting on it.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
Ball 1's density is 500 kg/m³. The tension on the rope holding Ball 2 is 41.15 N. The tension on the rope holding Ball 3 is 121.24 N.
Let's solve this problem step-by-step to better understand floating and submerged objects in water.
A.) Which is true for Ball 1?
Ball 1 floats with half of it exposed above the water level. This means the density of Ball 1 must be half the density of water. Since the density of water is 1000 kg/m³, the density of Ball 1 is:
500 kg/m³B.) What is the tension on the rope holding the second ball, in newtons?
Ball 2 has a density of 893 kg/m³ and is held below the surface of water. The buoyant force is equal to the weight of the volume of water displaced by Ball 2.
Calculate the volume of Ball 2: (Volume of a sphere = 4/3 π r³)
r = 21 cm r = 0.21 mVolume = (4/3) π (0.21)³ = 0.0388 m³Calculate the buoyant force:
Buoyant Force = Density of Water x Volume of Water x gBuoyant Force = 1000 kg/m³ x 0.0388 m³ x 9.8 m/s² Bouyant force = 380.24 NCalculate the weight of Ball 2:
Weight = Density of Ball 2 x Volume x gWeight = 893 kg/m³ x 0.0388 m³ x 9.8 m/s² = 339.09 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 339.09 N - 380.24 N = -41.15 N (negative indicates an upward force)Tension on the rope for Ball 2: 41.15 NC.) What is the tension on the rope holding the third ball in N?
Ball 3 has a density of 1320 kg/m³ and is also fully submerged, suspended by a rope.
Calculate the weight of Ball 3:
Weight = Density of Ball 3 x Volume x gWeight = 1320 kg/m³ x 0.0388 m³ x 9.8 m/s² = 501.48 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 501.48 N - 380.24 N = 121.24 NTension on the rope for Ball 3: 121.24 NThis example shows how to calculate buoyant forces and tensions for submerged and floating objects.
On a day that the temperature is 21.0°C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00 109 N/m2 and the compressive strength to be 2.00 109 N/m2. (The coefficient of linear expansion of concrete is 1.2 10-5(°C−1).) (a) What is the stress in the cement on a hot day of 33.0°C? N/m2
Answer:
1*10^6 N/m^2
Explanation:
Coefficient of Linear Expansion for Concrete = α = 1.2 x 10^-5 (°C)^-1
Change in temperature = ΔT
ΔT= T2 - T1
ΔT = 33 - 21
ΔT = 12°C
ΔL = α * L(i) * ΔT
ΔL = (1.2 x 10^-5 (°C)^-1) * L(i) * (12°C)
ΔL = 1.44 x 10^-4
Stress = F / A
Strain = ΔL / L
Strain = (1.44*10^-4) * (L) / L
Strain = 1.44*10^-4
Y = Stress / Strain
Stress = Y * Strain
Stress = (7.00*10^9 N/m^2) * (1.44*10^-4)
Stress = 1*10^6 N/m^2
Thus, the stress in the cement on a hot day of 33° is 1*10^6 N/m^2
1. Car Down Incline w Friction An automobile weighing 4250 lb starts from rest at point A on a 6o incline and coasts through a distance of 500 ft to point B. The brakes are then applied, causing the automobile to come to a stop at point C, 70 ft from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefficient of static friction between the tires and the road.
Answer:
Explanation:
Let θ be the inclination
downward acceleration on an inclined plane
= g sinθ
= 32 x sin6
a = 3.345 ft /s
a ) for knowing the speed at point B
v² = u² + 2 a s , v is final velocity , u is initial velocity , a is acceleration and s is distance travelled .
v² = 0 + 2 x 3.345 x 500
= 3345
v = 57.8 ft /s
from point B to C , the car decelerates so we shall find deceleration
v² = u² + 2 a s
0 = 3345 + 2 x a x 70 ( v becomes u here )
a = - 23.9 m /s²
net force on car during deceleration
= μmgcosθ - mg sinθ where μ is coefficient of static friction ,
= mg ( μcosθ - sinθ )
deceleration = g ( μcosθ - sinθ )
g ( μcosθ - sinθ ) = 23.9
( μcosθ - sinθ ) = .74
μcosθ = .74 + .104
= .8445
μ = .8445 / .9945
= .85 .
Final answer:
To calculate the maximum deceleration of a car heading down a 6° slope under different road conditions, we can use the coefficient of static friction. On dry concrete, the deceleration is approximately 2.12 m/s². On wet concrete, the deceleration is approximately 1.62 m/s². On ice, the deceleration is approximately 2.04 m/s².
Explanation:
To calculate the maximum deceleration of a car heading down a 6° slope, we need to consider the road conditions. Assuming the weight of the car is evenly distributed on all four tires, and that the tires are not allowed to slip during the deceleration, we can calculate the deceleration for different road conditions.
(a) On dry concrete, the coefficient of static friction can be calculated using the equation μs = tan(θ), where θ is the angle of the slope. In this case, the coefficient of static friction is approximately 0.105 and the maximum deceleration is approximately 2.12 m/s².
(b) On wet concrete, the coefficient of static friction is typically lower than on dry concrete. Let's assume a coefficient of 0.08. In this case, the maximum deceleration is approximately 1.62 m/s².
(c) On ice, assuming a coefficient of static friction of 0.100, the same as for shoes on ice, the maximum deceleration is approximately 2.04 m/s².
A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid that has a temperature of 40.0 degrees Celsius is poured over the glass, completely covering it, and the temperature at equilibrium is 57.0 degrees Celsius. The mass of the glass and the liquid is the same. Determine the specific heat capacity of the liquid
Answer:
741 J/kg°C
Explanation:
Given that
Initial temperature of glass, T(g) = 72° C
Specific heat capacity of glass, c(g) = 840 J/kg°C
Temperature of liquid, T(l)= 40° C
Final temperature, T(2) = 57° C
Specific heat capacity of the liquid, c(l) = ?
Using the relation
Heat gained by the liquid = Heat lost by the glass
m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)
Since their mass are the same, then
C(l)ΔT(l) = C(g)ΔT(g)
C(l) = C(g)ΔT(g) / ΔT(l)
C(l) = 840 * (72 - 57) / (57 - 40)
C(l) = 12600 / 17
C(l) = 741 J/kg°C
A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True False
Answer:
The question above is repeated twice.
Removing the repetition, we have: A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True or False?
The correct answer to the question is "False"
Explanation:
The question above, illustrates a phenomenon referred to as "Doppler effect"
The Doppler effect only changes the frequency of the sound which explains how wavelength changes when a wave source is moving toward or away from an object. The Doppler effect occurs when a source of waves and/or observer move relative to each other.
When a sound source is moving toward the observer (a person sitting in the plane) in the case above, the observer will hear a higher pitch as the source approaches. That is, the plane stands still on the ground and the siren is driven toward it.This is due to a decrease in the amplitude of the sound wave.
However, If the observer moves toward the stationary source, the observed frequency is higher than the source frequency. In this case, A plane flies toward a stationary siren.
λ = v/f = vT,
where T is the period,
The relationship between frequency, speed, and wavelength is:
f = v/λ
v represents the speed of sound through the medium.
Doppler effect depends on things moving, as the observer moves, the frequency becomes higher as the distance decreases. If the observer moves and the distance becomes larger, it means that the sound frequency becomes lower.
Final answer:
The Doppler effect explains why a person sitting in a plane moving toward or away from a stationary siren at 1/4 the speed of sound perceives the same frequency of sound. This phenomenon is true due to the relative motion between the source of sound and the observer.
Explanation:
True
The phenomenon described in the question is related to the Doppler effect in physics. When a source of sound and an observer are in motion relative to each other, the frequency of the sound waves changes due to this motion. In this case, when the plane is moving toward or away from the siren at 1/4 the speed of sound, the observer perceives the same frequency of sound from the siren.
A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the end of a table of height h2 , landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1 + h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.
Without using equations, explain why making h1 very small would cause d to be small, even though h2 would be large.
Without using equations, explain why making h2 very small would cause d to be small, even though h1 would be large
Derive an equation for d in terms of h1, h2, m, and physical constants, as appropriate.
Write the equation or step in your derivation in part (b) (not your final answer) that supports your reasoning in part (a)i.
Briefly explain your choice.
Write the equation or step in your derivation in part (b) (not your final answer) that supports your reasoning in part (a)ii.
Briefly explain your choice.
If the experiment is repeated on the Moon without changing h1 or h2 , will the new landing distance d be greater than, less than, or the same as the landing distance when the experiment is performed on Earth?
_____Greater than _____Less than _____The same as
The horizontal distance (d) a block travels after being released from a slide is limited by the initial velocity from the slide and the time of flight from the table. Mathematical considerations of kinetic energy and projectile motion demonstrate why the distances h1 and h2 are crucial factors. On the Moon, the block would travel further due to reduced gravity.
Explanation:The horizontal distance (d) that a block travels after sliding down a slide and falling off a table is dependent on both the vertical height dropped and the velocity with which it leaves the table. If we make height h1 (the slide) very small, the velocity of the block at the bottom of the slide and consequently at the end of the table would be small because it would have converted a smaller amount of potential energy into kinetic energy. This would result in a small horizontal distance (d) even though height h2 (the table) is large. Conversely, making height h2 (the table) very small would mean that the block doesn't have much height to fall from, which limits the total time it has to move horizontally, again resulting in a small d.
In analyzing both scenarios mathematically, the relationship between the horizontal distance and the heights would involve equations of motion and energy conservation. The step that supports reasoning in part (a)i would involve the equation for kinetic energy at the end of the slide (KE = 1/2 m[tex]v^2[/tex]), which is maximized when h1 is large. Similarly, the step supporting part (a)ii is Newton's equations of motion for projectile motion (particularly, time of flight = sqrt(2h2/g)), where increasing h2 increases the time the block spends in air and thus d.
When repeating the experiment on the Moon, the new landing distance d will be greater than the landing distance when performed on Earth. This is because the acceleration due to gravity on the Moon is less than on Earth, which increases the time the block spends in the air.
Calculate How much energy is transferred as useful energy
=A 98% efficient kettle that has a total input of 2000J
Answer:1960j
Explanation:
total input energy=2000j
98% of total input energy is useful
98% of 2000
98/100 x 2000
(98 x 2000) ➗ 100
196000 ➗ 100=1960
1960j is useful
The amount of useful transferred energy is 1960 J.
What is energy?A body's capacity for work is measured in terms of energy. It cannot be produced or eliminated. There are numerous types of energy, including thermal, electrical, fusion, electrical, and nuclear. Energy has the ability to change its forms.
What is efficiency?Efficiency is essentially a measurement of the amount of labour or energy that can be saved throughout a process. In other words, it's similar to comparing the energy input and output in any particular system. For instance, we observe that many processes result in the loss of effort or energy like vibration or waste heat.
Given parameters:
Total input energy; I =2000 Joule.
efficiency of the kettle; η = 98%.
We have to find useful output energy of the kettle: O = ?
We know that: output energy = efficiency × input energy
= 98% × 2000 J.
= 98/100 x 2000 J.
= (98 x 2000) ÷ 100 J.
= 196000 ÷ 100 J
= 1960 J.
Hence, the useful transferred energy is 1960 J.
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A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV. The beam is then entered into a region between two parallel metal plates with potential difference 120 V and a separation 8 mm, perpendicular to the direction of the field. What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates?
Answer:
The magnetic field required required for the beam not to be deflected is [tex]B = 0.0036T[/tex]
Explanation:
From the question we are told that
The charge on the particle is [tex]q = +2e[/tex]
The mass of the particle is [tex]m = 6.64 *10^{-27} kg[/tex]
The potential difference is [tex]V_a = 1.8 kV = 1.8 *10^{3} V[/tex]
The potential difference between the two parallel plate is [tex]V_b = 120 V[/tex]
The separation between the plate is [tex]d = 8 mm = \frac{8}{1000} = 8*10^{-3}m[/tex]
The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV
[tex]KE_b = PE_b[/tex]
Generelly
[tex]KE_b = \frac{1}{2} m v^2[/tex]
And [tex]PE_b = q V_a[/tex]
Equating this two formulas
[tex]\frac{1}{2} mv^2 = q V_a[/tex]
making v the subject
[tex]v = \sqrt{\frac{q V_a}{2 m} }[/tex]
Substituting value
[tex]v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }[/tex]
[tex]v = 41.65*10^4 m/s[/tex]
Generally the electric field between the plates is mathematically represented as
[tex]E = \frac{V_b}{d}[/tex]
Substituting value
[tex]E = \frac{120}{8*10^{-3}}[/tex]
[tex]E = 15 *10^3 NC^{-1}[/tex]
the magnetic field is mathematically evaluate
[tex]B = \frac{E}{v}[/tex]
[tex]B = \frac{15 *10^{3}}{41.65 *10^4}[/tex]
[tex]B = 0.0036T[/tex]
. Block m1 slides along a frictionless surface at speed v1 = 4 m/s. Then it undergoes a onedimensional elastic collision with stationary block m2 = 2m1. Next, block m2 undergoes a one-dimensional elastic collision with stationary block m3 = 2m2. (a) What is the speed of block m3? Are the (b) speed, (c) kinetic energy, and (d) momentum of block m3 greater than, less than, or the same as the initial values for m1?
Answer:
a) v3 = 1 m/s
c) K3 < K1
d) p3 = p1
Explanation:
a) To solve this problem you use the conservation of the linear momentum in elastic collision.
In the first case you have:
[tex]p_i=p_f\\\\m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
but the second block is at rest, then v2i = 0m/s:
[tex]m_1v_{1i}=m_1v_{1f}+m_2v_{2f}[/tex]
Furthermore, you can assume that the first object stops just after the collision with the second one. From this last expression you obtain the value of the second object:
[tex]v_{2f}=\frac{m_1v_{1i}}{m_2}\\\\m_2=2m_1\\\\v_{2f}=\frac{m_1(4m/s)}{2m_1}=2\ m/s[/tex]
Then, you use the conservation of momentum for the second case, in which the second objects impact the third one:
[tex]m_2v'_{2i}+m_3v_{3i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{3i}=0\\\\m_2v'_{2i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{2f}=0\\\\m_2v'_{2i}=m_3v_{3f}\\\\v_{3f}=\frac{m_2v'_{2i}}{m_3}[/tex]
where again it has assumed that the second object stops, just after the impact with the third object. v'_2i = v_2f (in order to distinguish). BY using the fact m3 = 2m2 you obtain:
[tex]v_{3f}=\frac{m_2(2m/s)}{2m_2}=1\ m/s[/tex]
Then, you obtain that v3 < v2 < v1
c) The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
you compute for all the three objects:
[tex]K_1=\frac{1}{2}m_1(4m/s)^2=8m_1\ m^2/s^2\\\\K_2=\frac{1}{2}m_2(2m/s)^2=\frac{1}{2}(2m_1)(4m^2/s^2)=4m_1\ m^2/s^2\\\\K_3=\frac{1}{2}m_3=(1m/s)^2=\frac{1}{2}(2m_2)(1\ m^2/s^2)=\frac{1}{2}(2(2m_1))(1 m^2/s^2)=2m_1\ m^2/s^2[/tex]
then, k3 < k2 < k1
d) For the momentum you have:
[tex]p_1=4m_1\ m/s\\\\p_2=m_2(2m/s)=(2m_1)(2m/s)=4m_1\ m/s\\\\p_3=m_3(1m/s)=(2m_2)(1m/s)=(2(2m_1))(1m/s)=4m_1\ m/s[/tex]
p1 = p2 = p3