Answer:
Explanation:
Gravitational Potential Energy at earth surface [tex]U_1=\frac{GM_em}{R_e}[/tex]
Gravitational Potential Energy at height h is [tex]U_2=\frac{GM_em}{R_e+h}[/tex]
Energy required to lift the satellite [tex]E_1=U_1-U_2[/tex]
[tex]E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}[/tex]
Now Energy required to orbit around the earth
[tex]E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}[/tex]
[tex]\Delta E=E_1-E_2[/tex]
[tex]\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}[/tex]
[tex]E_1=E_2[/tex] (given)
[tex]\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0[/tex]
[tex]\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0[/tex]
[tex]h=\frac{R_e}{2}[/tex]
[tex]h=3.19\times 10^6\ m[/tex]
(b)For greater height [tex]E_1[/tex] is greater than [tex]E_2[/tex]
thus energy to lift the satellite is more than orbiting around earth
Downstream peripheral pulses have a higher pulse pressure because the pressure wave travels faster than the blood itself. What occurs in peripheral arterial disease?
Answer:
Explanation: Peripherial arterial disease is a blood circulation disorder which occurs when the blood vessels outside the heart and brain to block or spasm. This is as a result of the pulse decreasing rather than increasing in amplitude.
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left sphere, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn. which answer is correct?
A) Both the spheres are neutral.
b) Left sphere is negatively charged, another is charged positively.
C) Right sphere is negatively charged, another is charged positively.
D) Both the spheres are charged positively.
E) Both the spheres are charged negatively
Answer: Option (C) is the correct answer.
Explanation:
As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.
Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.
As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.
Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.
During a snowball fight , your opponent distracts you by throwing a snowball at you in a high arc. she thriws snowballs with a speed of 24.5 m/s and the first is thrown at 70.0 degrees. As you are watching the first snowball, she throws a second at a lower angle. If both snowballs cover the same horizontal distance, at what angle should the second be thrown?
Answer:
Explanation:
Given
launch velocity [tex]u=24.5\ m/s[/tex]
first ball launch angle [tex]\theta _1=70^{\circ}[/tex]
Suppose another ball is thrown at an angle of [tex]\theta _2[/tex]
Both ball have same range
Range of Projectile [tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]R_1=\frac{u^2\sin 2\theta _1}{g}[/tex]
For second ball
[tex]R_2=\frac{u^2\sin 2\theta _2}{g}[/tex]
[tex]R_1=R_2[/tex]
[tex]\frac{u^2\sin 2\theta _1}{g}=\frac{u^2\sin 2\theta _2}{g}[/tex]
[tex]\sin 2\theta _1=\sin 2theta _2[/tex]
Either [tex]\theta _1=\theta _2[/tex] or
[tex]2\theta _1=180-2\theta _2[/tex]
I.e. [tex]\theta _2=90-\theta _1[/tex]
[tex]\theta _2=90-70=20^{\circ}[/tex]
so another ball must be thrown at [tex]20^{\circ}[/tex]
Lance arrives at the airport (with flowers and balloons in hand) to welcome a friend. Her plane is delayed. While waiting, he notices that it takes 2 minutes 47 seconds to get down the hall on the moving sidewalk. while walking (not on the moving sidewalk) it took him 112 seconds. If he walks while on the sidewalk, how long (s) will it take him?
Answer:
t = 67.04 s
Explanation:
given,
time taken by Lace on the side walk = 2 min 47 s
= 167 s
time taken by the Lace while walking = 112 s.
now, time taken by the Lace when he is walking on the side walk = ?
Assume the distance be equal to 'd'
speed of Lace on side walk
we know, distance = speed x time
[tex]v_{sidewalk}=\dfrac{d}{167}[/tex]
speed of the lace while walking
[tex]v_{walking}=\dfrac{d}{112}[/tex]
time taken by Lace to cover the distance by walking on the side walk
[tex]t = \dfrac{distance}{v_{sidewalk}+v_{walking}}[/tex]
[tex]t = \dfrac{d}{\dfrac{d}{167}+\dfrac{d}{112}}[/tex]
[tex]t = \dfrac{167\times 112}{167+112}[/tex]
t = 67.04 s
time taken by the Lace walking on the side walk is equal to 67.04 s
An object is dropped out of an airplane that is moving horizontally at 350 m/s and is 22,000 m above the ground. Ignoring friction, what will its approximate VF X be on impact?
Answer:
Vx = 350m/s
Vy = 656.99 m/s
Explanation:
at the time of drop, the horizontal velocity of the object is same as of plane.
Since friction is ignored, horizontal velocity remains unchanged.
Vx = 350m/s
but the Vy increases due to gravitational acceleration
V^2 - U^2 = 2as
V: FINAL VELOCITY
U: INITIAL VELOCITY
U = 0 m/s
[tex]v=\sqrt{2as}= \sqrt{2(9.81)(22000)} = 656.99 m/s[/tex]
The length of the assembly decreases by 0.006 in. when an axial a- force is applied by means of rigid end plates. Determine
(a) the magnitude of applied force,
(b) the corresponding stress in the steel core.
Final answer:
The stress in a steel rod due to an axial force is calculated based on the weight of the rod above the cross-section in question. Stress is calculated using σ = P / A, where P is the force and A is the cross-sectional area. The stress differs depending on the height from the lower end due to varying weights above the point.
Explanation:
The question is discussing a scenario in which a cylindrical steel rod undergoes axial deformation due to an applied force. This deformation could be either tension (elongation) or compression (shortening). The rod is experiencing stress and strain as a reaction to this force. To solve for the magnitude of the force and the stress in the steel core, we use principles from mechanical engineering, specifically statics and material science.
To find the normal stress in the steel rod at different heights when the rod is fastened vertically, we need to calculate the weight of the portion of the rod above the point in question. This is because the weight causes a compressive stress in the steel rod. The stress (σ) at a given point can be calculated using the formula σ = P / A, where P is the force (weight) acting on the cross-section and A is the area of the cross-section.
(a) For the cross-section 1.0 m from the lower end, only half of the rod's weight is above this point, and thus contributing to the stress.
(b) For the cross-section 1.5 m from the lower end, the portion of the rod above this point is 0.5 m, so only the weight of this segment contributes to the stress.
The names of the seven days of the week are derived from the names of the bodies of the solar system that are visible to the naked eye.
Answer:
Baja naja kanaka kanaka man's kana kana
An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys be pulled in order to lift the object a distance of 1.4 m?
Answer:
4.2 m
Explanation:
Note: If energy is conserved, i.e no work is done against friction
Work input = work output.
Work output = Force output × distance,
Work input = force input × distance moved moved.
Therefore,
input force×distance moved = output force × distance moved........................Equation 1
Given: input force = 80 N, output force = 240 N, output distance = 1.4 m
Let input distance = d
Substitute into equation 1
80×d = 240×1.4
80d = 336
d = 336/80
d = 4.2 m.
Thus the rope around the pulley must be pulled 4.2 m
What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? (Assume that the range of visible wavelengths of light in air is 380 - 760 nm.)
Answer:
1315789.47368 lines/m
Explanation:
m = Order = 1
[tex]sin\theta[/tex] = 1 For maximum condition
[tex]\lambda[/tex] = Wavelength = 760 nm maximum
From Rayleigh criteria we have the expression for the gap
[tex]d=\dfrac{m\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1\times 760}{1}\\\Rightarrow d=760\ nm[/tex]
The number of lines is the reciprocal of the slit distance
[tex]n=\dfrac{1}{d}\\\Rightarrow n=\dfrac{1}{760\times 10^{-9}}\\\Rightarrow n=1315789.47368\ /m[/tex]
The number of lines is 1315789.47368 per meter
Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m. Which statement accurately describes something that Yolanda can do as a part of her study?a. She can use destructive interference to generate a wave with an amplitude of 3.1 m. b. She can use constructive interference to generate a wave with an amplitude of 1.5 m. c. She can use destructive interference to generate a wave with an amplitude of 2.8 m. d. She can use constructive interference to generate a wave with an amplitude of 3.5 m.
statement accurately describes something that Yolanda can do as a part of her study She can use destructive interference to generate a wave with an amplitude of 2.8 m.
The correct option is (c).
To determine the resulting amplitude of waves under interference, we consider the principles of constructive and destructive interference.
1. Constructive Interference: When two waves meet crest to crest or trough to trough, they add up to produce a wave with an amplitude equal to the sum of the individual amplitudes.
For Yolanda's waves:
[tex]\[ \text{Amplitude} = 2 \, \text{m} + 3 \, \text{m} = 5 \, \text{m} \][/tex]
Option b. She can use constructive interference to generate a wave with an amplitude of 1.5 m is incorrect because the resulting amplitude from constructive interference must be greater than or equal to the sum of the individual amplitudes.
2. Destructive Interference: When two waves meet crest to trough, they cancel each other out partially or completely. The resulting amplitude is the absolute difference between the amplitudes of the individual waves.
For Yolanda's waves:
[tex]\[ \text{Amplitude} = |2 \, \text{m} - 3 \, \text{m}| = |-1 \, \text{m}| = 1 \, \text{m} \][/tex]
Option a. She can use destructive interference to generate a wave with an amplitude of 3.1 m is incorrect because the resulting amplitude from destructive interference must be less than or equal to the difference between the individual amplitudes.
Option d. She can use constructive interference to generate a wave with an amplitude of 3.5 m is incorrect because the resulting amplitude from constructive interference must be greater than or equal to the sum of the individual amplitudes.
Therefore, option c. She can use destructive interference to generate a wave with an amplitude of 2.8 m is the accurate statement.
A ball is dropped off of a tall building and falls for 6 seconds before landing on the ground. Consider how far the ball falls in its first 3 seconds of free fall (from t = 0 s to t = 3 s) compared to how far it falls in its next 3 seconds (from t = 3 s to t = 6 s).
Answer:
In the last six seconds the ball will fall 3 times the distance it fell in the first 3 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
Total distance
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s_1=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m[/tex]
At t = 3 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s_1=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s_2=44.145\ m[/tex]
In the next 3 seconds it will fall
[tex]s_2=176.58-44.145=132.435\ m[/tex]
Dividing the equations
[tex]\dfrac{s_2}{s_1}=\dfrac{132.435}{44.145}\\\Rightarrow s_2=3s_1[/tex]
In the last six seconds the ball will fall 3 times the distance it fell in the first 3 seconds
A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground.
Answer:
[tex]\displaystyle \theta' =0.24\ rad/s[/tex]
Explanation:
Rate Of Change
Let some variable y depend on time t. we can express y as a function of t as
[tex]y=f(t)[/tex]
The instant rate of change of y respect to t is the first derivative, i.e.
[tex]y'=f'(t)[/tex]
The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula
[tex]\displaystyle tan\theta =\frac{y}{x}[/tex]
Since x is constant, we take the derivative with respect to time by using the chain rule:
[tex]\displaystyle sec^2\theta \ \theta' =\frac{y'}{x}[/tex]
Solving for [tex]\theta'[/tex]
[tex]\displaystyle \theta' =\frac{y'}{xsec^2\theta}[/tex]
Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet
[tex]\displaystyle tan\theta =\frac{y}{x}[/tex]
[tex]\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}[/tex]
Knowing that
[tex]\sec^2\theta=1+tan^2\theta[/tex]
[tex]\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2[/tex]
[tex]\displaystyle \sec^2\theta=\frac{25}{16}[/tex]
The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:
[tex]\displaystyle \theta' =\frac{8}{12\ \frac{25}{16}}[/tex]
[tex]\displaystyle \theta' =\frac{32}{75}\ rad/s[/tex]
[tex]\boxed{\displaystyle \theta' =0.43\ rad/s}[/tex]
The magnitude of vector A is 15.0 units and points in the direction 330° counterclockwise from the positive x-axis. Calculate the x- and y-components of this vector.
Answer:
[tex]V_{x}[/tex] = 12.99 , [tex]V_{y}[/tex] = -7.5
Explanation:
v=15units , θ=330°
the magnitude of the horizontal and vertical components are gotten by the formula below
[tex]V_{x}[/tex] = Vcosθ
[tex]V_{y}[/tex] = Vsinθ
[tex]V_{x}[/tex] = 15*cos330
[tex]V_{x}[/tex] = 12.99 (positive implies to the right)
[tex]V_{y}[/tex] =15*sin330
[tex]V_{y}[/tex] = -7.5 (negative implies downwards)
A roller coaster is climbing up the highest hill on its track. At which point will the front car of the coaster most likely have its greatest potential energy?
Answer:
top of the highest hill
Explanation:
The potential energy of an object is given by
[tex]U=mgh[/tex]
where,
m = Mass of the object
g = Acceleration due to gravity
h = Perpendicular distance from the ground.
It can be seen that the potential energy is proportional to height
[tex]U\propto h[/tex]
Here, the car will have maximum potential energy when it is at the maximum distance from the ground.
Hence, the car will be at the top of the highest hill
The front car of a roller coaster will have its greatest potential energy at the highest hill due to the stored gravitational potential energy that can be converted into kinetic energy as it descends.
Potential energy in a roller coaster system is greatest at the highest hill. At this point, the roller coaster's gravitational potential energy is at its maximum as it has the most stored energy that can be converted into kinetic energy as it descends. This transformation between potential and kinetic energy explains why the roller coaster's front car will most likely have its greatest potential energy at the top of the highest hill.
You need to make a spring scale to measure the mass of objects hung from it. You want each 2.1 cm length along the scale to correspond to a mass difference of 0.10 kg. What should be the value of the spring constant?
Answer:
46.67 N/m
Explanation:
mass, m = 0.1 kg
distance, y = 2.1 cm = 0.021 m
Let K be the spring constant.
F = mg = Ky
0.1 x 9.8 = K x 0.021
K = 46.67 N/m
Thus, the spring constant of the spring is 46.67 N/m.
Answer:
k = 46.67 N/m
Explanation:
given,
mass, m = 0.10 Kg
Length, x = 2.1 cm = 0.021 m
spring constant, k = ?
we know,
spring force
F = k x
and also
F = m g
computing both the equation
k x = m g
[tex]k = \dfrac{mg}{x}[/tex]
[tex]k = \dfrac{0.1\times 9.8}{0.021}[/tex]
k = 46.67 N/m
Spring constant for the spring is equal to k = 46.67 N/m
When designing a user interface, the most important information should be placed in the ______ of the screen.
Answer:
upper-left corner
Explanation:
Most vital information are positioned in a place where users can view them clearly and without obstruction.
A bobsled team pushes a 132-kg bobsled. If the combined push of the team is 450.0 N and the bobsled also experiences a force of friction of 35 N, what is the acceleration of the bobsled?
Answer:
3.14 m/s^2 is the acceleration
Explanation:
You must use the formula Fnet=ma
So the total forces acting on the system are the 450N push force and the 35N friction force, which acts in the opposite direction.
Thus, the sum of the forces is 450 -35= 415N
Also you know the mass is 132 kg, so to find the acceleration divide the previously calculated net force by the mass
So 415/132 =3.14 m/s^2
Hope this helped:)
The bobsled with a mass of 132 kg and a net force of 415 N after factoring in friction accelerates at about 3.14 meters per second squared.
To determine the acceleration of the bobsled, we will use Newton's second law, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. The net force in this scenario is the combined push minus the force of friction. Therefore, the equation to use is:
Net force = Total push - Force of friction
Net force = 450.0 N - 35 N = 415 N
Now, we use this net force to find the acceleration:
Acceleration (a) = Net force / Mass
a = 415 N / 132 kg
a ≈ 3.14 m/s2
The bobsled will accelerate at approximately 3.14 meters per second squared when the net force acting on it is 415 N.
Let's say you have two tuning forks which are supposed to produce the same frequency, 512 Hz. One is of good quality, but the other is cheaply made and vibrates at 510 Hz. If you bonk them both how many beats per second will you hear
Answer:
= 2 beats per seconds
Explanation:
From |f -f'| = modulus of the difference between the frequency given.f = 510Hz and f' = 512HzDifference between the frequency will give us the number of beat per seconds.i.e 2 beats per secondsThese also shows how to get the period of the tuning forks.
As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the average force exerted on the car that hits a line of water barrels and takes 1.3 s to stop.
Answer:
-[tex]29.2\times 10^{3} N[/tex]
Explanation:
We are given that
Mass of cars= m=1900 kg
Initial speed of car=u=20 m/s
Final speed of car=v=0
Time=[tex]\Delta t[/tex]=1.3 s
We have to find the average force exerted on the car.
Average force=[tex]\frac{change\;in\;momentum}{\Delta t}[/tex]
[tex]F_{avg}=\frac{mv-mu}{1.3}[/tex]
[tex]F_{avg}=\frac{1900(0)-1900(20)}{1.3}[/tex]
[tex]F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N[/tex]
Hence, the average force exerted on the car that hits a line of water barrels=-[tex]29.2\times 10^{3} N[/tex]
Final answer:
Using the change in velocity and time, the acceleration of the car is calculated, followed by the application of Newton's second law to find the average force exerted on the car as 29,222 N.
Explanation:
To determine the average force exerted on the car that hits a line of water barrels, we can use the equations of motion and Newton's second law of motion. The car goes from 20 m/s to rest, which means that its change in velocity is -20 m/s. Given the time of 1.3 seconds for this change, we can find the acceleration using the formula:
a = Δv / t
Where Δv is the change in velocity and t is the time taken. The acceleration a is therefore:
a = -20 m/s / 1.3 s = -15.38 m/s²
Now, using Newton's second law of motion, Force (F) = mass (m) × acceleration (a), we can find the force:
F = 1900 kg × -15.38 m/s² = -29222 N
The negative sign indicates that the force is in the opposite direction to the motion of the car. Thus, the average force exerted on the car is 29,222 N.
Final answer:
Using the change in velocity and time, the acceleration of the car is calculated, followed by the application of Newton's second law to find the average force exerted on the car as 29,222 N.
Explanation:
To determine the average force exerted on the car that hits a line of water barrels, we can use the equations of motion and Newton's second law of motion. The car goes from 20 m/s to rest, which means that its change in velocity is -20 m/s. Given the time of 1.3 seconds for this change, we can find the acceleration using the formula:
a = Δv / t
Where Δv is the change in velocity and t is the time taken. The acceleration a is therefore:
a = -20 m/s / 1.3 s = -15.38 m/s²
Now, using Newton's second law of motion, Force (F) = mass (m) × acceleration (a), we can find the force:
F = 1900 kg × -15.38 m/s² = -29222 N
The negative sign indicates that the force is in the opposite direction to the motion of the car. Thus, the average force exerted on the car is 29,222 N.
Solar energy breaks oxygen molecules apart in the stratosphere, releasing ____, which combine(s) with additional oxygen molecules to form ____.
Answer:
Oxygen atom
Ozone Molecule
Explanation:
The upper stratosphere and Mesosphere consist of the ozone layer. This layer helps in protecting us from the UV radiations of the Sun. This layer converts the UV radiation into heat energy.
The UV radiation breaks down the Oxygen molecules in oxygen atom which further combines with oxygen molecule to form Ozone. This Ozone molecule breaks to form Oxygen atom and Oxygen molecule. Oxygen atom further recombines with Oxygen molecule to make Ozone molecule. This is known as Ozone-Oxygen cycle. The excess energy is released as heat.
Everyday around 400 million metric tons of Ozone is produced through this cycle.
Malleability, because it shows how adjacent layers of positive ions can move relative to one another while remaining in full contact with the electron sea.
Answer:
Explanation: Metals with close-packed structures such as copper, gold, silver, zinc, magnesium, etc. are more malleable than those with the bcc structure (tungsten, vanadium, chromium, etc.).
In the close-packed structure, the planes can slip past each other relatively easily. In the bcc structure, there are no close-packed planes, and much greater corrugation between atoms at different levels. This makes it much harder for one row to slide past another.
The carrying capacity of an environment for a particular species at a particular time is determined by the a. number of individuals in the species. b. distribution of the population. c. reproductive potential of the species. d. supply of the most limited resources.
Answer:
d. supply of the most limited resources.
Explanation:
the carrying capacity of a species in an environment is determined by the number of individual in an environment that the limited resources can sustain indefinitely without fighting for food, vegetation, water and light to sustained themselves.
the carrying capacity can be determined by how much the limited resources in the environment can sustain the species. the resources needed by species in a particular environment varies from habitat to habitat.
for some species, the limiting resources needed is food which consequently affect the way they reproduced and affect the population size.
The carrying capacity of an environment for a specific species is defined by the availability of the most limited resources, and it indicates the maximum population size that the environment can sustain.
Explanation:The carrying capacity of an environment for a particular species at a particular time is determined by the supply of the most limited resources. This refers to the maximum population size of a species that an environment can sustain indefinitely, given the availability of resources such as food, water, and space. It's not solely determined by the number of individuals in the species, the distribution of the population, or the reproductive potential of the species, rather it is driven by the resource which is in shortest supply.
Learn more about Carrying Capacity here:
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Doug has filled a long tray with an even layer of sand. Doug then uses two bricks to prop up one side of the tray. Next, Doug uses a watering can to pour water onto the high side of the tray. what natural process is he making
Answer:
Percolation of water through the soil.
Explanation:
When a tray containing a mixture of dough and sand is propped on one end of the tray with bricks then there is a slope from one end of the tray to the other end of the tray.
When water is poured on the higher end of the tray then the it tries to flow from the higher altitude to the lower altitude of the tray by the process of percolation through the granular spaces created due to the mixing of sand in the dough. Sand being granular in structure is easily permeable to water and passes water easily through its tiny pores.
Similar is the process of percolation of water through the layers of soil on the earth which also contains granular particles in them.
A -1.0@mC charge experiences a 10ni@N electric force in a certain electric field. What force would a proton experience in the same field?
Answer: A proton would experience the same magnitude of force.
Explanation: The equation relating Force F charge Q and Electric Field E, is given as
E = F/Q
The magnitude of the force felt would be the same since + and - charge as same magnitude from the question (-1C and +1C). But the direction of force of the proton which as a positive charge would be towards/same direction as the Electric Field.
Imagine you are on a spaceship that is under constant acceleration of 1g. If the spaceship was stationary at the start, approximately how many miles will you travel in one year?
Answer:
3035281543986.976 mi
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration = g
Time taken
[tex]t=1\ y\\\Rightarrow t=365.25\times 24\times 60\times 60[/tex]
u = Initial velocity = 0
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times (365.25\times 24\times 60\times 60)^2\\\Rightarrow s=4.8848\times 10^{15}\ m[/tex]
Converting to mi
[tex]1\ m=\dfrac{1}{1609.34}\ mi[/tex]
[tex]4.8848\times 10^{15}\ m=4.8848\times 10^{15}\times \dfrac{1}{1609.34}\ mi\\ =3035281543986.976\ mi[/tex]
Number of miles travelled in one year is 3035281543986.976 mi
Give an example of an isometric contraction?
Explanation:
A classic instance of an isometric contraction would be to bear an object before you. The object's weight will be pulling down, but the movement of the object would be opposed by your arms and hands with equivalent force pushing up the object.
Assuming that your arms do not lift or fall, your biceps will contract isometrically. Isometric contraction generate force without changing the length of the muscle.
Answer:
Pushing a wall without bending or motion in the arm joints.
Explanation:
Isometric literally means that the measurement of a particular thing does not change.
Here isometric contraction is related with the contraction of the muscles without the movements of the joints. During the process the muscles feel the load and muscle fibers contract without change in length.
Every contraction of muscles in the skull is isometric.
You are hungry and decide to go to your favorite neighborhood fast-food restaurant. You leave your apartment and take the elevator 10 flights down (each flight is 3.0 m) and then go 15 m south to the apartment exit. You then proceed 0.2 km east, turn north, and go 0.1 km to the entrance of the restaurant.
Part A
Determine the displacement from your apartment to the restaurant. Use unit vector notation for your answer, being sure to make clear your choice of coordinates.
Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction, and the z-axis vertical.
Part B
How far did you travel along the path you took from your apartment to the restaurant?
Part C
What is the magnitude of the displacement you calculated in part
Answer:
A) R = (200 i ^ + 100 j ^ + 30k ^) m , B) L = 223.61 m , C) R = 225.61 m
Explanation:
Part A
This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.
Let's write the displacements
Descending from the apartment
10 flights of 3 m each, the total descent is 30 m
Z = 30 k ^ m
Offset at street level
L1 = 0.2 i ^ km
L2 = 0.1 j ^ km
Let's reduce everything to the SI system
L1 = 0.2 * 1000 = 200 i ^ m
L2 = 100 j ^ m
The distance traveled is
R = (200 i ^ + 100 j ^ + 30k ^) m
Part B
The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane
L² = x² + y²
L = √ (200² + 100²)
L = 223.61 m
Part C
The magnitude of travel, let's use the Pythagorean theorem for the sum
R² = x² + y² + z²
R = √ (30² + 200² + 100²)
R = 225.61 m
Analysis of an air sample reveals that it contains 3.5 x 10-6 g/L of carbon monoxide. Express the concentration of carbon monoxide in lb/ft3. (Use 1.00 lb = 454 g; 1 in = 2.54 cm)
To convert the concentration of carbon monoxide from g/L to lb/ft3, we need to perform two unit conversions: from grams to pounds and from liters to cubic feet. The concentration of carbon monoxide in lb/ft3 is approximately 2.7248 x 10-10 lb/ft3.
Explanation:To convert the concentration of carbon monoxide from g/L to lb/ft3, we can use unit conversion factors. First, we need to convert grams to pounds using the given conversion factor 1.00 lb = 454 g. Then, we can convert liters to cubic feet by using the conversion factor 1 L = 0.0353 ft3. Let's perform the calculations:
Therefore, the concentration of carbon monoxide in lb/ft3 is approximately 2.7248 x 10-10 lb/ft3.
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Older railroad tracks in the U.S. are made of 12-m-long pieces of steel. When the tracks are laid, gaps are left between sections to prevent buckling when the steel termally expands. If a track is laid at 16*C, how large should the gaps be if the track is not to buckle when the temperature is as high as 50*C?
Answer: ∆L = 0.49cm ≈ 0.50cm
Therefore there should be 0.5 cm gap between each piece of steel.
Explanation:
Thermal expansion of steel is the increase in size of steel as a result of increased temperature. It can be represented by the mathematical expression:
∆L = L(k)∆T .....1
Where;
∆L is the change in length
L is the initial length
∆T is the change in temperature
k is the specific Linear expansion coefficient.
Given;
L = 12m
∆T = 50°C - 16°C = 34°C
k (for steel) = 1.2 × 10^-6 /C
Substituting the values into the equation 1
∆L = 12 × 34 × 12×10^-6
∆L = 4896 × 10^-6 m
∆L = 0.49cm ≈ 0.50cm
Therefore there should be 0.5 cm gap between each piece of steel.
There should be a gap of 0.5 cm between each of the piece of steel.
Based on the given information,
• The initial length of the tracks is 12 m.
• The laying of the tracks was done at the temperature of 16 °C (Initial temperature), the final temperature is 50 °C.
• The increase in the size of the steel as an outcome of enhanced temperature is known as the thermal expansion of steel.
The mathematical representation of thermal expansion is,
ΔL = L(k)ΔT
• Here ΔL refers to the change in length, L is the initial length, k is the specific linear expansion, which is 1.2 × 10⁻⁶/C for steel, and ΔT is the change in temperature, which is 50 °C - 16 °C = 34 °C.
Now putting the values we get,
ΔL = 12 × 34 × 12 × 10⁻⁶
ΔL = 4896 × 10⁻⁶ m
ΔL = 0.49 cm or 0.5 cm
Thus, the gap between each piece of steel should be 0.5 cm.
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A chemist dissolves 377.mg of pure potassium hydroxide in enough water to make up 130.mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25°C.) Round your answer to 3 significant decimal places.
Answer:
The answer to your question is pH = 1.36
Explanation:
Data
Mass = 377 mg of KOH
Volume = 130 ml
pH = ?
pH = -log[H⁺¹]
Process
1.- Calculate the moles of KOH
molecular mass of KOH = 39 + 16 + 1 = 56 g
56 g of KOH --------------- 1 mol of KOH
0.377 g KOH ------------- x
x = (0.377 x 1) / 56
x = 0.0057 moles
2.- Calculate the concentration
Molarity = number of moles / volume
Molarity = 0.0057 / 0.13
Molarity = 0.044
3.- Calculate the pH of the solution
pH = - log [0.044]
pH = 1.36