The ratio between rotational energy and translational kinetic energy is [tex]1.47\cdot 10^{-3}[/tex]
Explanation:
The translational kinetic energy of the ball is given by:
[tex]KE_t = \frac{1}{2}mv^2[/tex]
where
m is the mass of the ball
v is the speed of the ball
The rotational kinetic energy of the ball is given by
[tex]KE_r = \frac{1}{2}I\omega^2[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular speed
The moment of inertia of a solid sphere through its axis is given by
[tex]I=\frac{2}{5}mR^2[/tex]
where
m is the mass of the ball
R is its radius
Substituting into the previous equation,
[tex]KE_r = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2 \omega^2[/tex]
The ratio between the two energies is
[tex]\frac{KE_r}{KE_t}=\frac{\frac{1}{5}mR^2 \omega^2}{\frac{1}{2}mv^2}=\frac{2R^2 \omega^2}{5v^2}[/tex]
And substituting:
R = 3.91 cm = 0.0391 m
v = 33.6 m/s
[tex]\omega=52.1 rad/s[/tex]
we find:
[tex]ratio = \frac{2(0.0391)^2(52.1)^2}{5(33.6)^2}=1.47\cdot 10^{-3}[/tex]
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The ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere should be [tex]1.47.10^-3[/tex].
Calculation of the ratio:Since
The translational kinetic energy of the ball should be
[tex]KE_t = 1/2mv^2[/tex]
here
m is the mass of the ball
v is the speed of the ball
Now
The rotational kinetic energy of the ball should be
[tex]KE_r = 1/2Iw^2[/tex]
Here
I is the moment of inertia
w is the angular speed
Now
The moment of inertia of a solid sphere through its axis should be
[tex]I = 2/5mR^2[/tex]
Here
m is the mass of the ball
R is its radius
So,
[tex]KE_r = 1/2(2mR^2)w^2\\\\= 1/5mR62w^2[/tex]
Now
the ratio be
[tex]KE_r/KE_t = (1/2mR^2w^2) / (1/2mv^2)\\\\= (2R^2 w^2) / (5v^2)\\\\= 290.0391)^2 (52.1)^2 / 5(33.6)^2\\\\= 1.47*10^-3[/tex]
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A particle's velocity is described by the function vx=kt2m/s, where vx is in m/s, t is in s, and k is a constant.
The particle's position at t0=0s is x0 = -8.10 m . At t1 = 2.00 s , the particle is at x1 = 8.00 m .
Determine the units of k in terms of m and s.
Answer:
ms^(-3)
Explanation:
V = kt^2
m/s = k (s)^2
k = m/ s^(3)
k = ms^(-3)
Imagine you can take all the atoms in a single drop of water and put them on a single line as closely packed as they can be. How long would that line be in meters? There is approximately 1022 atoms in a droplet of water.A. 1012 meters which is bigger than the distance between Sun and Earth.B. 1020 meters, the size of a galaxy.C. 103 meters, this is one kilometer.D. 107 meters which is about the circumference of the Earth.
Answer:
A. 10^12 meters which is bigger than the distance between Sun and Earth
Explanation:
In a case where we take all the atoms in a single drop of water and put them on a single line as closely packed as they can be, the total length of the line would be a function of the diameter of an atom.
Total length L = Diameter of an atom d × number of atom in a droplet of water N
L = dN
N = 10^22
d ~= 0.1nm = 10^-10m
L = 10^22 × 10^-10 m
L = 10^12 m
So, the length would be approximately 10^12 m
distance between the sun and earth is 147.34million km
D= 1.47 × 10^11 m
L > D
Therefore, the length would be approximately 10^12 m
Which is greater than the distance between the earth and the sun.
Two identical conducting small spheres are placed with their centers 0.290 m apart. One is given a charge of 12.0 nC and the other a charge of -21.0 nC.Find the electric force exerted by one sphere on the other.
Answer:
Force between two identical conducting ball will be [tex]2.65\times 10^{-5}N[/tex]
Explanation:
We have given there is two identical conducting small spheres having charge [tex]q_1=12nC=12\times 10^{-9}C[/tex] and [tex]q_2=-21nC=-21\times 10^{-9}C[/tex]
Distance between centers of the ball is 0.290 m
So r = 0.290 m
We have to fond the electric force between the balls
According to Coulomb's law force between two charges is given by
[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]
So force [tex]F=\frac{9\times 10^{9}\times 12\times 10^{-9}\times 21\times 10^{-9}}{0.293^2}=2.65\times 10^{-5}N[/tex]
So force between two identical conducting ball will be [tex]2.65\times 10^{-5}N[/tex]
In an experiment to create a pendulum, each member of the group measured the length of the string to be a slightly different value. How do we describe these small differences when reporting the length?
Answer:Kobe
Explanation:
The small difference at the time of reporting the length should be the random uncertainty.
The following information related to the random certainty is:
It arises at the time when an experiment should be repeated and less changed arises.It can occur because of the measurement techniques or the design of the experiment. Also, if the length varies from the real one that varies for the various members so it is a random certainty.There should be a difference in the length as compared to the real one for various members.
Therefore we can conclude that The small difference at the time of reporting the length should be the random uncertainty.
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a- what constant acceleration, in SI units, must a car have to gofrom zero to 60 mph in 10 s?
b- what fraction of g is this?
c- how far has the car traveled when it reaches 60 mph? Giveyour answer both in SI units and in feet
Answer:
(a) 0.017m/s^2
(b) 17/100,000
(c) 0.17m, 0.558ft
Explanation:
(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s
Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2
(b) g = 9.8m/s^2, a = 0.0017m/s^2
a/g = 0.0017/9.8 = 0.00017 = 17/100,000
(c) Distance = speed × time = 0.017m/s × 10s = 0.17m
Distance in foot = 0.17 × 3.2808ft = 0.558ft
Answer:
a) acceleration a = 2.68 m/s^2
b) fraction of g = 0.273 or 27.3%
c) distance travelled d = 134m (in SI unit)
d = 439.5ft (in feet)
Explanation:
a) converting 60mph (miles per hour) to m/s
∆v = 60 miles/hour × 1609.344meters/mile × 1/3600 seconds/hour
∆v = 26.82m/s
t = 10s
acceleration = change in velocity/time = ∆v/t
Acceleration a = 26.82/10 = 2.68m/s^2
b) acceleration due to gravity is g = 9.8m/s^2
Fraction of g = a/g = 2.68/9.8 = 0.273
Fraction = 0.273 or 27.3%
c) distance travelled can be calculated using the equation below.
d = ut + 0.5at^2 ....1
Initial speed u = 0
time t = 10s
Acceleration a = 2.68m/s^2
d = distance covered.
From equation 1, putting u = 0
d = 0.5at^2
d = 0.5(2.68 × 10^2)
d = 134m
Converting meters to feet
d = 134m × 3.28ft/m
d = 439.5ft
What is the mass of the star?You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8×107m and its rotation period to be 22.3 hours . You have previously determined that the planet orbits 2.2×1011m from its star with a period of 422 earth days. Once on the surface you find that the free-fall acceleration is 12.2m/s2.
Answer:
M=4.736×10³⁰kg
Explanation:
Given Data
Diameter of planet= 1.86×10⁷m
Radius R=Diameter/2=1.86×10⁷/2⇒0.93×10⁷m
Period of rotation of planet t=22.3 hr⇒80280 seconds
Period of revolution around star T=422 earth days⇒3.646×10⁷ seconds
Radius of orbit r=2.2×10¹¹m
Acceleration of gravity on surface of planet g=12.2 m/s²
To Find
Mass of star
Solution
To find mass of star.For that we will consider the revolution of planet around the star with given period
[tex]T^{2}=\frac{4\pi ^{2}r^{3} }{GM}\\ M=\frac{4\pi ^{2}r^{3} }{GT^{2}}\\M=\frac{4\pi ^{2}(2.2*10^{11} )^{3} }{(6.67*10^{-11} )(3.646*10^{7} )^{2}}\\M=4.736*10^{30} kg[/tex]
An electric field can induce an electric dipole in a neutral molecule (or atom) by pushing the positive and negative charges inside the molecule in opposite directions. The dipole moment of the induced dipole is directly proportional to the electric field at the molecule. That is, p=αE , where p is the induced dipole moment, α is called the polarizability of the molecule, and E⃗ is the electric field at the molecule. A stronger electric field at the molecule results in a more polarized molecule and causes a larger dipole moment p . A molecule with a larger polarizability, α, will be more polarized when subjected to the same electric field. Find an expression for the magnitude of the force F⃗ ionondipole.Express your answer in terms of the variables q, r, α and appropriate constants.
The force on an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The magnitude of the force can be expressed as (1/(4πε0)) * (2αEext)/(r3) * q, where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule.
Explanation:The force experienced by an ion interacting with an induced dipole can be found using Coulomb's law. The force can be represented as Fionondipole = qEinduced, where q is the charge of the ion and Einduced is the induced electric field. The induced electric field can be expressed as Einduced = (1/(4πε0)) * (2αEext)/(r3), where α is the polarizability of the molecule, Eext is the external electric field, and r is the distance between the ion and the molecule. Combining these expressions, the magnitude of the force Fionondipole can be written as:
Fionondipole = (1/(4πε0)) * (2αEext)/(r3) * q
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Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(≈22,000miles).What is the period of a satellite in a geosynchronous orbit?
Answer:
T = approximately 24 hs.
Explanation:
In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.
Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:
Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)
By definition, ω =ΔФ / Δt
For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),
so we can replace ω by (2*π/T), solving then for T:
T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).
The period of a satellite in a geosynchronous orbit is 24 hours, matching the Earth's rotation. The satellite follows a circular orbit keeping it in the same place relative to Earth's surface. Satellites in this orbit provide various services due to their continuous presence above a particular location.
Explanation:The period of a satellite in a geosynchronous orbit is 24 hours. It's called geosynchronous because it matches the Earth's rotation period which is also 24 hours, this is how they are able to stay over the same point on the Earth's surface. The satellites move in elliptical orbits, but those in geosynchronous orbit follow a circular path so they maintain a constant altitude and remain in constant position relative to the Earth's surface.
As there are many satellites launched each year, being in a geosynchronous orbit helps these satellites to provide continuous services like weather tracking, communications, global positioning systems, etc. These satellites also avoid being captured by Earth's gravity and pulled into the atmosphere because they are at a distance of approximately 36000km from the earth's surface, where gravity is much weaker.
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A car starts from rest, speeds up with constant acceleration, and travels 350 meters in 5 seconds. What was the average velocity of the car in m/s? 70 m/s omputer's answer now shown above.
What was the final velocity of the car (in m/s)?_______________ m/s Submit Answer Tries 0/2 Consider that the car started at rest.
What was the change in velocity of the car (in m/s)?________ m/s Submit Answer Tries 0/2
What was the acceleration of the car (in m/s2)? _______m/s2
Remember that the acceleration is the change in velocity divided by the change in time
Answer
70 m/s, 140 m/s, 28 m/s²
Explanation:
Average velocity = total distance travel / time = 350 / 5 = 70 m/s
average velocity = (initial velocity + final velocity) / 2
70 × 2 = 140 m/s
change in velocity = final velocity - initial velocity = 140 m/s - 0 = 140 m/s
acceleration = (final velocity - initial velocity) / t = 140 m/s / 5 = 28 m/s²
Answer:
1. Final velocity is 140m/s
2. Change in velocity is 140m/s
3. Acceleration is28m/s^2
Explanation:
Initial velocity u=0,
Distance travelled S=350m
Time taken t=5 seconds
The car move constant acceleration, then we can use any of equations of motion
v=u+at
v^2=u^2+2as
s=ut+(at^2)/2
Using equation 3
S=ut+(at^2)/2
350= 0×5+ (a ×5^2)/2
350= 0+(a×25)/2
350=25a/2
350×2=25a
700=25a
a=700/25
a=28m/s^2. Answer
1. Now, using equation 1
v=u+at
v=0+28×5
v=0+140
v=140m/s. Answer
2. The changed in the velocity of the car is final velocity minus the initial velocity
Change in velocity = v-u
Change in velocity = 140-0
Therefore,
Change in velocity is 140m/s. Answer
3. The acceleration has been answered before solving and it is
a= 28m/s^2
In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the distance in units of centimeter and you know the unit equality 1inch = 2.54 centimeter. By which conversion factor will you multiply 28.4 inch in order to perform the unit conversion?
To solve this problem we will apply the concept designed to generate the conversion from one unit to another. Basically, what is sought is to eliminate the units of the denominator and the numerator of the conversion factor and the unit to be converted respectively, leaving the units of the new unit. In mathematical terms this is,
[tex]1 in = 2.54cm[/tex]
If we want to convert 28.4in then the conversion factor versus the unit would be
[tex]x = 28.4 in (\frac{2.54cm}{1in})[/tex]
[tex]x = 72.136 cm[/tex]
Therefore the factor of conversion will be 2.54 and the final units for the value given is 72.136cm
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter.
Answer:
0.556m
Explanation:
unit conversion
58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m
[tex]7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3[/tex]
Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula
[tex]V_h = \frac{4}{3}\pi r^3[/tex]
The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part
[tex]V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)[/tex]
Let water density [tex]\rho_w = 1000 kg/m^3[/tex], then the buoyant force is the weight of water displaced by the shell
[tex]F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)[/tex]
For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell
[tex]F_b = W_s[/tex]
[tex]g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s[/tex]
[tex]g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)[/tex]
[tex]1000*0.0246 = 7870(0.0246 - r^3)[/tex]
[tex]24.6 = 193.602 - 7870r^3[/tex]
[tex]r^3 = (193.602 - 24.6)/7870 = 0.02147[/tex]
[tex]r = \sqrt[3]{0.02147} = 0.278 m[/tex]
So the inner diameter is 0.278*2 = 0.556 m
To find the inner diameter of the hollow sphere, one must take steps of calculating the total volume of the sphere, then the weight, the displaced water volume and finally the total internal volume. With this information, the inner diameter of the sphere can be worked out.
Explanation:To solve this physics problem, we need to find the inner diameter of a hollow spherical iron shell that is floating almost completely submerged in water. The student has provided the outer diameter of the shell (58.2 cm) and the density of iron (7.87 g/cm3). Using these values, we'll use some concepts from physics, specifically principles of buoyancy and formulas for volume and density.
The first step will be to calculate the total volume of the sphere using the outer diameter. This can be done with the formula for the volume of a sphere: V=(4/3)πR^3, where R is the radius, which is half the diameter.
Since the sphere is floating almost completely submerged, it's displacing a volume of water equivalent to its own weight. Now that we know the sphere's volume, we can calculate the sphere's weight using the density of iron.
The weight then can be used to calculate the volume of water displaced, which leads us to the total internal volume of the sphere (total volume - displaced volume). Then the inner diameter can be found by rearranging the formula of the volume of the sphere.
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An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B with a take-off velocity of 270 km/h, determine (a) the acceleration a, (b) distance AB.
Answer:
a)The acceleration of the airplane is 2.5 m/s².
b)The distance AB is 1125 m.
Explanation:
Hi there!
a)The equation of velocity of an object moving in a straight line with constant acceleration is the following:
v = v0 + a · t
Where:
v = velocity of the object at time "t".
v0 = initial velocity.
a = acceleration.
t = time
We have the following information:
The airplane starts with zero velocity (v0 = 0) and its velocity after 30 s is 270 km/h (converted into m/s: 270 km/h · 1000 m/1 km · 1 h /3600 s = 75 m/s). Then, we can solve the equation to obtain the acceleration:
75 m/s = a · 30 s
75 m/s / 30 s = a
a = 2.5 m/s²
The acceleration of the airplane is 2.5 m/s².
b)The distance AB can be calculated using the equation of position of an object moving in a straight line with constant acceleration:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time "t".
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
If we place the origin of the frame of reference at A, then, x0 = 0. Since the airplane is initially at rest, v0 = 0. So, the equation gets reduced to this:
x = 1/2 · a · t²
Let´s find the position of the airplane after 30 s:
x = 1/2 · 2.5 m/s² · (30 s)²
x = 1125 m
The position of B is 1125 m away from A (the origin), then, the distance AB is 1125 m.
(a) the acceleration of the plane is 2.5 m/s²
(b) The distance AB is 1125m
Equations of motion:
(a) Given that the initial velocity of the plane at point A is u = 0
and the final velocity of the plane at point B of take-off is v = 270 km/h
[tex]v=\frac{270\times1000}{3600}m/s=75m/s[/tex]
the time taken to teach point B is t = 30s
So from the first equation of motion, we get:
[tex]v=u+at[/tex]
where a is the acceleration.
[tex]75=a\times30\\\\a=2.5\;m/s^2[/tex]
(b) From the second equation of motion we get,
[tex]s=ut+\frac{1}{2} at^2[/tex]
where s is the distance
[tex]s = \frac{1}{2}\times2.5\times30^2\\\\s=1125m[/tex]
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A spring with a spring constant of 23.6 N/m has a mass attached that exerts a force of 6.28 Newtons. What is the displacement?
Answer:
26.6 cm
Explanation:
We are given that
Spring constant=23.6 N/m
Force exert=6.28 N
We have to find the displacement.
We know that Hooke's law
[tex]F=kx[/tex]
Where k= Spring constant
x=Displacement
Using the formula
Then, we get
[tex]6.28=23.6 x[/tex]
[tex]x=\frac{6.28}{23.6}=0.266m[/tex]
We know that 1 m=100 cm
[tex]x=0.266\times 100=26.6 cm[/tex]
Hence, the displacement =26.6 cm
Determine the values of mm and nn when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,000,000,000 kgkg. Enter mm and nn, separated by commas.
Final answer:
Scientific notation is essential in handling large numbers in physics. The gravitational constant is crucial for calculating gravitational forces. Understanding Earth's mass involves utilizing known values such as its radius and the gravitational constant.
Explanation:
Scientific Notation: Scientific notation is commonly used in astronomy and other sciences to represent very large or small numbers efficiently.
Gravitational Constant: The gravitational constant, denoted by G, is a fundamental constant used to calculate gravitational forces.
Earth's Mass Calculation: Earth's mass, represented as 5.97 x 10²⁴ kg, can be calculated using known values like its radius and the gravitational constant.
The United States possesses the ten largest warships in the world, aircraft carriers of the Nimitz class. Suppose one of the ships bobs up to float 11 cm higher in the ocean water when 50 fighters take off from it in a time interval of 25 min, at a location where the free-fall acceleration is 9.78 m/s^2. The planes have an average laden mass of 29,000 kg. Find the horizontal area enclosed by the waterline of the ship.
Answer:The horizontal area enclosed by the waterline of the ship is 12,797.88m2
Explanation: Mass of the ship=29000kg
Total mass of 50 ships= 50×29000
Total Mass=1450000kg
Volume of water with mass= mass/ density
Assuming density of ocean water =1030kgm3
Volume=1450000/1030
Volume=1407.77m3
Area enclosed by the ship =A=V/h
h=11cm=0.11m
Area= 1407.77/0.11
Area=12,797.88m2
Suppose we have no idea what the voltages of our batteries are. What must these voltages be if a 100 amp current ?ows uniformly across all three loops of the circuit above?
Answer:hello
Explanation:
Assume your sampling frequency (fs) is 500 Hz. If you want to capture the 50Hz signal for 5 seconds, calculate the number of samples (N) at the assumed fs to get the desired 5 seconds of data.
Answer:
N = 2500
Explanation:
By definition, the sampling frequency (fs) is:
[tex] f_{s} = N / T [/tex]
where N: is the number of samples and T: is the time
Hence, the number of samples at the sampling frequency, 500 Hz, to get 5 seconds of data is:
[tex] N = f_{s} \cdot T = 500 s^{-1} \cdot 5 s = 2500 [/tex]
I hope it helps you!
To capture a 50Hz signal for 5 seconds with a sampling frequency of 500 Hz, we need 2500 samples.
Explanation:To capture the 50Hz signal for 5 seconds, we need to calculate the number of samples (N) at the assumed sampling frequency (fs) of 500 Hz. According to the Nyquist theorem, the highest frequency we can accurately monitor is half the sampling frequency. In this case, the Nyquist frequency would be 250 Hz. Since the signal frequency is 50 Hz, it is well below the Nyquist frequency. Therefore, we need to sample the signal at a rate greater than twice the signal frequency, which is 100 samples per second. Multiplying this by the desired duration of 5 seconds, we get the total number of samples required:
N = fs * duration = 500 Hz * 5 s = 2500 samples
Expand (1-2tz t^2)^-1/2 in powers of t assuming that t is small. Collect the coefficients of t0 , t1 . and t2 .
Answer:
Coefficients are 1/2, -z and +1
Explanation:
The step by step explanation is in the attachment given below.
A record for stair climbing was achieved by a man who raced up the 1600 steps of the Empire State Building to a height of 320 m in 10 min 59 sec. If his mass was 75 kg, what power did he generate (in both watts and horsepower) while climbing?
Answer:
Power will be 356.90 watt
Explanation:
We have given total number of steps in the stair = 1600
And height id the stair h = 320 m
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Mass is given m = 75 kg
So work done in climbing the stairs [tex]W=mgh=75\times 9.8\times 320=235200J[/tex]
Time is given t = 10 min 59 sec
So time [tex]t=10\times 60+59=659sec[/tex]
We know that power is rate of doing work
So power [tex]P=\frac{Work}{time}=\frac{235200}{659}=356.90watt[/tex]
So power will be 356.90 watt
The man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.
To calculate the power generated by the man while climbing the stairs, we can use the work-energy principle. The work done by the man in climbing to a height against the force of gravity is equal to the change in his gravitational potential energy. The power is then the work done divided by the time taken to do it.
The gravitational potential energy (PE) is given by:
[tex]\[ PE = mgh \][/tex]
where:
-[tex]\( m \)[/tex] is the mass of the man (75 kg),
-[tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \) on Earth),[/tex]
- [tex]\( h \)[/tex] is the height climbed (320 m).
First, we calculate the work done (PE):
[tex]\[ PE = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 320 \, \text{m} \][/tex]
[tex]\[ PE = 235,440 \, \text{J} \][/tex]
Now, we convert the time from minutes to seconds because power is typically calculated in joules per second (watts). The time taken is 10 minutes and 59 seconds, which is:
[tex]\[ 10 \times 60 \, \text{s} + 59 \, \text{s} = 659 \, \text{s} \][/tex]
Power (P) is the work done divided by the time taken:
[tex]\[ P = \frac{PE}{t} \][/tex]
[tex]\[ P = \frac{235,440 \, \text{J}}{659 \, \text{s}} \][/tex]
[tex]\[ P \approx 357.26 \, \text{W} \][/tex]
To convert watts to horsepower, we use the conversion factor [tex]\( 1 \, \text{hp} = 746 \, \text{W} \):[/tex]
[tex]\[ P_{\text{hp}} = \frac{P}{746} \][/tex]
[tex]\[ P_{\text{hp}} \approx \frac{357.26 \, \text{W}}{746} \][/tex]
[tex]\[ P_{\text{hp}} \approx 0.479 \, \text{hp} \][/tex]
Therefore, the man generated approximately 357.26 watts or 0.479 horsepower while climbing the stairs.
The final answer in the correct format is:
[tex]\[ \boxed{357.26 \, \text{W}} \][/tex]
[tex]\[ \boxed{0.479 \, \text{hp}} \][/tex]
The answer is:[tex]0.479 \, \text{hp}.[/tex]
The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for this particle
Answer:
f=qB/2[tex]\pi[/tex]m
Explanation:
The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for this particle
solution
The Lorentz force [tex]F_{lorentz} =F_{centripetal}[/tex]
is the centripetal force
and makes the particles path to revolve in a circle:
qvB=mv^2/r
radius=r
m=mas of the particle
B=magnetic flux
q=quantity of charge
v=velocity of the particle
v=qBr/m
where v is the velocity of the particle
recall the velocity v=rω
v=2*Pi*f*r
[tex]2\pi *f*r[/tex]=qBr/m
f=qB/2[tex]\pi[/tex]m
the cyclotron frequency is therefore f=qB/2[tex]\pi[/tex]m
the period(the time it take to make a complete oscillation) will be the inverse of frequency
T=2[tex]\pi[/tex]m/qB
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge 1.07 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.
(a) What is the x component of the electric force exerted by A on C?
N
(b) What is the y component of the force exerted by A on C?
N
(c) Find the magnitude of the force exerted by B on C.
N
(d) Calculate the x component of the force exerted by B on C.
N
(e) Calculate the y component of the force exerted by B on C.
N
(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C.
N
(g) Similarly, find the y component of the resultant force vector acting on C.
N
(h) Find the magnitude and direction of the resultant electric force acting on C.
Answer:
a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N
h) 24.3 N θ = 44.2º
Explanation:
a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.
So, Fcax = 0
b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.
Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²
Fyca = 29. 9 N
c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:
rbc² = (3.00 m)² + (4.00m)² = 25.0 m²
⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²
⇒ Fbc = 21.7 N
d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:
Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:
cos θ = x/r = 4.00 / 5.00 m =
Fcbx = 21.7*(-0.8) = -17.4 N
e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:
Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N
f) The sum of both x components gives :
Fcx = 0 + (-17.4 N) = -17.4 N
g) The sum of both y components gives :
Fcy = 29.9 N + (-13.0 N) = 16.9 N
h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:
Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²[tex]\sqrt{((17.4)^{2} +(16.9)^{2}}[/tex] = 24.3 N
The angle from the horizontal can be found as follows:
Ф = arc tg (16.9 / 17.4) = 44.2º
DC versus AC problem. Suppose your DC power supply is set to 15 V and the vertical setting on the oscilloscope is at 5 V/div.
How many divisions will the trace deflect if you connect the output of the power supply to the scope with the "A input" switch set on DC?
How many divisions if the switch is on AC?
Answer:
DC = 3 div. AC= 0
Explanation:
When the input is directed to the input circuitry, and the "A input" switch is set to DC, the pure 15 V DC signal will be showed on the screen, so, if the vertical setting is at 5 V /div, the trace will deflect exactly 3 div.
If the switch is on AC, as this setting inserts a capacitor in series (which is located here to block any unwanted DC component superimposed to an AC signal) the DC signal will be blocked, so no trace will be deflected on the screen (after completed the transient period).
In this case, there are 3 divisions if the switch is on direct current (DC), whereas there are 0 divisions if the switch is on alternating current (AC).
Alternating current (AC) and direct current (DC) are electrical circuits that may be able to carry different currents.A DC is an electrical circuit composed of any combination of constant voltage, constant currents, and resistors.An AC circuit is an electrical circuit powered by an alternating source, that may be current or voltage.In conclusion, there are 3 divisions if the switch is on direct current, whereas there are 0 divisions if the switch is on alternating current.
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Workmen are trying to free an SUV stuck in the mud. To extricate the vehicle, they use three horizontal ropes, producing the force vectors shown in the figure.(Figure 1) Take F1 = 914N ,F2 = 704N , and F3 = 399N . Part A Find the x components of each of the three pulls. Enter your answer as three numbers, separated with commas. F1x,F2x,F3x = N SubmitMy AnswersGive Up Try Again Part B Find the y components of each of the three pulls. Enter your answer as three numbers, separated with commas. F1y,F2y,F3y = N SubmitMy AnswersGive Up Try Again Part C Use the components to find the magnitude of the resultant of the three pulls. F? = N SubmitMy AnswersGive Up Try Again Part D Use the components to find the direction of the resultant of the three pulls. Enter your answer as the angle counted from +x axis in the counterclockwise direction.
Answer:
Fx = (783.451 , 373.063 , 240.124 , 240.124) N
Fy = (470.745 , 597.026 , 318.66)
Fres =768.216
Theta = 77.2 degrees from + x
Explanation:
F1 = 914N; F2 = 704N ; F3 = 399N
Part A
Fx1 = F1*cos (A) = 914*cos (31) = +783.451 N
Fx2 = F2*sin (B) = 704*sin (32) = -373.063 N
Fx3 = F3*cos (C) = 399*cos (53) = -240.124 N
Sum of Fx = +170.26 N
Part B
Fy1 = F1*sin (A) = 914*sin (31) = +470.745 N
Fy2 = F2*cos (B) = 704*cos (32) = +597.026 N
Fy3 = F3*sin (C) = 399*sin (53) = -318.66 N
Sum of Fy = + 749.111 N
Part C
Fres = sqrt (Fx^2 + Fy^2 )
Fres = sqrt (170.26^2 + 749.111^2)
Fres = 768.216 N
Part D
Angle = arctan (Fy/Fx)
Angle = arctan (749.111/170.26)
Angle = 77.2 degrees
To find the x and y components of each pull, we use trigonometry. The magnitude of the resultant pull is found by summing up the squares of the x and y components and taking the square root. The direction of the resultant pull is determined using the inverse tangent function.
Explanation:To find the x components of each pull, we can use trigonometry. The x component of a force is given by the magnitude of the force multiplied by the cosine of the angle it makes with the x-axis. So, the x components of the three pulls are: F1x = F1 × cos(theta1), F2x = F2 × cos(theta2), F3x = F3 × cos(theta3).
To find the y components of each pull, we use the same approach but with the sine function. So, the y components of the three pulls are: F1y = F1 × sin(theta1), F2y = F2 × sin(theta2), F3y = F3 × sin(theta3).
Once we have the x and y components of each pull, we can find the magnitude of the resultant pull by summing up the squares of the x and y components and taking the square root: F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²).
To find the direction of the resultant pull, we can use the inverse tangent function. The angle counted from the +x axis in the counterclockwise direction is given by: theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).
So, the answers to the question are: F1x, F2x, F3x = 914N × cos(theta1), 704N × cos(theta2), 399N × cos(theta3); F1y, F2y, F3y = 914N × sin(theta1), 704N × sin(theta2), 399N × sin(theta3); F = √((F1x)² + (F2x)² + (F3x)² + (F1y)² + (F2y)² + (F3y)²); theta = atan(F1y + F2y + F3y / F1x + F2x + F3x).
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Find the magnitude of the net electric force exerted on a charge Q, located at the center of the square, for the following arrangement of charge: the two positive charges are on the top corners, and the two negative charges are on the bottom corners.
Express your answer in terms of the variables q, Q, a, and appropriate constants.
Explanation of finding the net electric force on a charge Q using Coulomb's Law in an arrangement with charges on the square's corners.
Explanation:To find the net electric force on a charge Q located at the center of a square with charges on its corners, we must consider the forces exerted by each charge using Coulomb's Law.
The net electric force on charge Q can be calculated by adding the individual force vectors from the charges Q₁ and Q₂.
By determining the force vectors and summing them up, we can express the net electric force on charge Q in terms of q, Q, a, and appropriate constants.
The magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]
To find the magnitude of the net electric force exerted on the charge Q at the center of the square, we can calculate the electric force exerted by each individual charge and then sum them up vectorially.
Let q be the magnitude of each corner charge and a be the side length of the square.
The electric force between two charges q separated by a distance r is given by Coulomb's law:
[tex]\[F = \frac{k \cdot q \cdot Q}{r^2}\][/tex]
Considering symmetry, the horizontal components of the electric forces from the charges on the left and right sides cancel out, as do the vertical components from the top and bottom charges.
Thus, the net electric force acting on Q is the sum of the forces from the top and bottom charges, which are equal in magnitude:
[tex]\[F_{\text{net}} = 2 \cdot F_{\text{top}}\][/tex]
[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{(\frac{a}{\sqrt{2}})^2}\][/tex]
[tex]\[F_{\text{net}} = 2 \cdot \frac{k \cdot q \cdot Q}{\frac{a^2}{2}}\][/tex]
[tex]\[F_{\text{net}} = \frac{4kqQ}{a^2}\][/tex]
Therefore, the magnitude of the net electric force exerted on the charge Q at the center of the square is [tex]\(\frac{4kqQ}{a^2}\).[/tex]
The Henry's law constant for CO2 is 3.6 ✕ 10−2 M/atm at 25°C. What pressure (in atm) of carbon dioxide is needed to maintain a CO2 concentration of 0.19 M in a can of lemon-lime soda?
To solve this problem we will use Henry's law. This law states that at a constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid. Mathematically it is formulated as follows:
[tex]C = K_H*P[/tex]
Where,
[tex]K_H[/tex] = Henry's constant for C02 at 25°C is equal to [tex]3.6*10^{-2}M/atm[/tex]
C = Gas concentration is 0.19M
Replacing we have,
[tex]0.19 M = (3.6*10^-2 M/atm)*P[/tex]
[tex]P = 5.277 atm[/tex]
Therefore the pressure of carbon dioxide is 5.277 atm
To calculate the pressure of CO2 needed to maintain a 0.19 M concentration in a can of soda, you apply Henry's law, which results in the pressure equalling 0.00684 atm.
Explanation:The subject of this question is Henry's law, which is used in Chemistry to relate the solubility of a gas in a liquid to the pressure of that gas above the liquid. In this case, we're asked to calculate the pressure of the CO2 gas needed to maintain a certain concentration in a can of soda. Henry's law is defined as: P = kH × C, where P is the gas pressure, kH is Henry's law constant, and C is the concentration.
To solve this problem, we rearrange Henry's law to find the pressure: P = kH × C. Given that the Henry's law constant for CO2 is 3.6 × 10−2 M/atm and the CO2 concentration is 0.19 M, we can apply these values and get P = (3.6 × 10−2 M/atm) × (0.19 M) which equals 0.00684 atm.
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A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table.
(a) Calculate the maximum speed of the object.
(b) Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the left as negative.
Answer:
1.137278672 m/s
+5.9 cm or -5.9 cm
Explanation:
A = Amplitude = 6.25 cm
m = Mass of object = 225 g
k = Spring constant = 74.5 N/m
Maximum speed is given by
[tex]v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s[/tex]
The maximum speed of the object is 1.137278672 m/s
Velocity is at any instant is given by
[tex]\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m[/tex]
The locations are +5.9 cm or -5.9 cm
Final answer:
The maximum speed of the object is approximately 1.06 m/s. When the object's velocity is one-third of the maximum speed, the object is located approximately 0.060 m from the equilibrium position.
Explanation:
To calculate the maximum speed of the object, the equation for the period of oscillation can be used. The period, T, can be calculated using the formula T = 2π√(m/k), where m is the mass of the object and k is the force constant of the spring. Substituting the given values, we have T = 2π√(0.225/74.5) = 0.942 s. The maximum speed, vmax, can be determined using the formula vmax = 2πA/T, where A is the amplitude of the oscillation. In this case, the amplitude is given as 6.25 cm, which goes to 0.0625 m. Substituting the values, we have vmax = 2π(0.0625)/0.942 = 1.06 m/s.
To find the locations of the object when its velocity is one-third of the maximum speed, we can use the equation for the displacement of an object undergoing simple harmonic motion. The equation is given as x = Acos(ωt + φ), where x is the displacement, A is the amplitude, t is the time, ω is the angular frequency, and φ is the phase constant. Since the object is released from rest, the phase constant is 0. The angular frequency can be calculated using the formula ω = 2π/T, where T is the period. Substituting the given values, we have ω = 2π/0.942 = 6.69 rad/s. The time when the velocity is one-third of the maximum speed can be found by rearranging the formula for velocity, v = ωAsin(ωt + φ), to T/6.69 = Asin(ωt + φ). Solving for t, we find that t ≈ 0.303 s. Substituting this value into the equation for displacement, we have x = Acos(ωt + φ) = 0.0625cos(6.69(0.303)) ≈ 0.060 m.
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 ADetermine 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 BHow far did you travel along the path you took from your apartment to the restaurant?Part CWhat is the magnitude of the displacement you calculated in part A?
Answer:
A. (200i + 85j - 30k)m
B. 345m
C. 219.37 m
Explanation:
UP: +VE Z
NORTH: +VE Y
EAST: +VE X
A.
DISPLACEMENT : (0.2*1000 i + (-15+0.1*1000)j + (-3*10)k) m
= (200i + 85j - 30k)m
B. total distance = 10*3+15+200+100 = 345m
C.
Magnitude = [tex]\sqrt{{200^{2} } +85^{2} +(-30)^{2}[/tex] = 219.37 m
unit vector = [tex]\frac{200i+85j-30k}{219.4}[/tex]
The displacement from the apartment to the restaurant is given by [tex]\Delta x =200\hat{i}+85\hat{j}-30\hat{k}[/tex].
The distance from the apartment to the restaurant is 345 m
The magnitude of the displacement from the apartment to the restaurant is 219.37 m.
Given that;
East is chosen as the X direction,North is chosen as the Y direction,Vertical is chosen as Z direction.Part A
The initial position can be identified as the origin, ie; [tex]x_i= 0 \hat{i}+0 \hat{j}+0 \hat{k}[/tex]To find the final position, let us find the position in each direction;
In X direction, [tex]x_1 = 0.2 \,km= 0.2 m = 200 \hat {i}[/tex]In Y direction, [tex]y_1 = -15\,m +0.1\,km=-15\,m+100\,m = 85 \,m = 85\hat{j}[/tex]In Z direction, [tex]z_1 = 3\times -10\,m = -30\,m = -30\hat{k}[/tex]Therefore, the final position is [tex]x_f = x_1+y_1+z_1=200\hat{i}+85\hat{j}-30\hat{k}[/tex]We know that displacement is the change in position.ie; displacement [tex]\Delta x = x_f - x_i =200\hat{i}+85\hat{j}-30\hat{k}[/tex]Part B
Here, we are asked to find the total distance covered.
For that, we add all the distances irrespective of the direction.
Distance (d) = vertical distance + south distance + east distance + north distance.[tex]d=(3\times 10 m)+ 15\,m + 200\,m + 100\,m = 345\,m[/tex]Part C
To find the magnitude of the displacement,
we know that the magnitude of a vector [tex]a\hat{i}+b\hat{j}+c\hat{k}[/tex] is [tex]\sqrt{(a^2 +b^2 + c^2)}[/tex]Therefore the magnitude of displacement, [tex]|\Delta x|= \sqrt{(200^2+85^2+30^2)} =219.37\,m[/tex]Learn more about distance and displacement here:
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What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?
To find the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, one must use the formula for energy density related to electric field strength, considering the permittivity of free space.
Explanation:To solve for the electric field strength that would store 12.5 J of energy in every 6.00 mm3 of space, we need to use the relationship between the electric field (E), the energy density (u), and the permittivity of free space (ε0). The energy density in an electric field is given by u = ½ε0E2. Rearranging for E, and substituting in the given values allows us to solve for the electric field strength.
Assuming ε0 = 8.85 x 10-12 C2/N·m2, and converting the volume from mm3 to m3 (6.00 mm3 = 6.00 x 10-9 m3), we can substitute these values into the rearranged formula E = √(2u/ε0) to find the electric field strength needed to store 12.5 J of energy.
Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is held 1.4 meters above the floor. Use the terms KE of GPE as your answers.When the ball is held motionless above the floor, the ball possesses only __________ energy.If the ball is dropped, its________ energy decreases as it falls.If the ball is dropped, its________ energy increases as it falls.
Answer:
When the ball is held motionless above the floor, the ball possesses only GPE energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.
Explanation:
If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.
If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.
[tex]K = \frac{1}{2}mv^2\\U = mgh[/tex]
Answer:
Explanation:
A. GPE
B. GPE
C. KE
A 5.0kg rock whose density is 4800 kg/m3 is suspended bya string such that half of the rock's volume is under water. What is the tension in the string?
Answer:
The tension in the string is 43.9 N.
Explanation:
Given that,
Mass of rock = 5.0 kg
Density of rock = 4800 kg/m³
We need to calculate the volume of rock
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
[tex]V=\dfrac{m}{\rho}[/tex]
Put the value into the formula
[tex]V=\dfrac{5.0}{4800}[/tex]
[tex]V=0.001041\ m^3[/tex]
We need to calculate the volume of water
[tex]V_{w}=\dfrac{V}{2}[/tex]
Put the value of volume
[tex]V_{w}=\dfrac{0.001041}{2}[/tex]
[tex]V_{w}=0.0005205\ m^3[/tex]
We need to calculate the mass of water displaced
Using formula of mass
[tex]m = 1000\times0.0005205[/tex]
[tex]m=0.5205\ kg[/tex]
We need to calculate weight of water displaced
Using formula of weight of water
[tex]W=0.5205\times9.8[/tex]
[tex]W=5.1009\ N[/tex]
Weight of rock is
[tex]W_{r}=5.0\times9.8[/tex]
[tex]W_{r}=49\ N[/tex]
We need to calculate the tension in the string
Using formula of tension
[tex]T=\text{weight of rock - weight of water displaced}[/tex]
Put the value into the formula
[tex]T=49-5.1009[/tex]
[tex]T=43.9\ N[/tex]
Hence, The tension in the string is 43.9 N.
The tension in the string is mathematically given as
T=43.9 N
What is the tension in the string?Question Parameter(s):
A 5.0kg rock whose density is 4800 kg/m3 suspended bya string
Generally, the equation for the Volume is mathematically given as
[tex]V=\frac{m}{\rho}[/tex]
Therefore
V=5.04/800
V=0.001041 m^3
for water
Vw=0.0005205 m^3
Weight of water
W=0.520*9.8
W=5.1009 N
Weight of rock
Wr=49N
In conclusion, Tension on string
T=49-5.1009
T=43.9 N
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