Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The surface temperature is about 37 K.
(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?
(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).
(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)
Answers
Answer:
a) The number density is 3.623 × 10⁻³ [tex]\frac{mol}{m^{3} }[/tex]
The mass of the atmosphere is 1.3 × 10²²Kg
b) The pressure is 10⁻²⁰ Millimeter of mercury
c) The mass mixing ratio is 0.0107
The partial pressure of ethane is 0.01114 Pa
Yes it is condensable because it boiling point is -88.5 C which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere is 37 K.
Explanation:
The explanation is on the first and second uploaded image
Related Questions
A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until their first American winter. Explain.
Answers
Explanation:
Water in the air (humidity) helps to dissipate static charge that builds up. If the air is very dry, the charge can't dissipate, so it builds up until there is enough to spark.
Tropical countries are typically more humid than the United States, but I guess that depends on where you are in the US.
A student's first experience with static electricity in an American winter is due to the dry air, which allows for greater accumulation and discharge of electrical charges, unlike in humid tropical climates.
This is because static electricity is more prevalent in cold, dry environments.
In tropical countries, the air tends to have higher humidity, which allows electrical charges to dissipate more easily. Moist air is a better conductor of electricity, thereby reducing the likelihood of a significant buildup of static charge.In contrast, during an American winter, the air is typically cold and dry. Dry air is a poor conductor, meaning that electrical charges are more likely to accumulate on surfaces and objects, leading to more frequent static electricity shocks. When you walk on a rug or take off a woollen sweater in such conditions, electrons can be transferred between your body and the surfaces, creating an imbalance.As a result, when you touch a metal object like a doorknob, the built-up charge is suddenly released, creating a spark and a shock. Even though the spark may carry a few hundred watts of power, it happens so quickly and involves such a small amount of current that it doesn't cause injury. The brief duration and low current ensure that the shock is felt but not harmful.In summary, the primary reason a student may experience static electricity for the first time in an American winter is due to the dry air conditions that favour the accumulation and sudden discharge of electric charges.
Make a prediction on how two A-tapes interact. Will they attract, repel or exert no force on each other?
Answers
Answer:
The two tapes will either attract or repel each other, depending on the nature of interaction. This is explained below:
Explanation:
When you rip the two pieces of tape off the table, there is a tug-of-war for electric charges between tape and table. The tape either steals negative charges (electrons) from the table or leaves some of its own negative charges behind, depending on what the table is made of (a positive charge doesn’t move in this situation). In any case, both pieces of tape end up with the same kind of charge, either positive or negative. Since like charges repel, the pieces of tape repel each other.
When the tape sandwich is pulled apart, one piece rips negative charges from the other. One piece of tape therefore has extra negative charges. The other piece, which has lost some negative charge, now has an overall positive charge. Because opposite charges attract, the two pieces of tape attract each other.
34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then load up the pickup and pump up itstires so that its total weight increases by 42% whilethe coefficient of rolling friction decreases by19%.
a) Now what horizontal force will you need to move the pickupalong the same road at the same speed? The speed is low enough thatyou can ignore air resistance.
Answers
Answer:
[tex]F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
[tex]F_H_n=F_f\\F_h=mg*u[/tex]
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
[tex]200=mg*u[/tex] Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
[tex]F_H=(1.42m*g*.81u)[/tex] Eq (2)
Divide Eq(2) and Eq (1)
[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N
A 8-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, determine the maximum volume flow rate of water.
Answers
The maximum Volume Flow Rate of water the pump can provide, given an efficiency of 82% and an elevation of 15 m, is approximately 0.033 L/s.
First, we must convert the pump's horsepower to a more usable unit in this context - like watts. In physics, 1 horsepower equals roughly 746 watts. Therefore, the pump has power of 8*746 = 5968 watts.
Given the mechanical efficiency (ME) and the height (h), the maximum work the pump can do is given by M.E. * Power. So, the pump does work of 0.82*5968 = 4895.76 watts.
The work done on the water by the pump is equal to the change in potential energy of the water, PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15 m). With rearranging, you could express m = Power/(g * h). But we're looking for the volume flow rate, not the mass flow rate, so we need to convert mass to volume. Since the density of water (ρ) is 1 kg/L, the volume flow rate = m/ρ = Power/(g * h * ρ).
Substituting all known values, we get: Volume flow rate = 4895.76 W /(9.8 m/s^2 * 15 m * 1 kg/L) = 0.033 L/s.
For more such questions on Volume Flow Rate, click on:
https://brainly.com/question/35117926
#SPJ3
There are great similarities between electric and gravitational fields. A room can be electrically shielded so that there are no electric fields in the room by surrounding it with a conductor. Can a room be gravitationally shielded? Explain.
Answers
Answer:
Can a room be gravitationally shielded? No, it can't.
Explanation:
the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.
In an automobile collision, how does an airbag lessen the blow to the passenger? Assume as a result of the collision, the passenger stops.a.The air bag decreases the momentum change of the passenger in the collision.b.
During the collision, the force from the air bag isgreater than would be the force from the wind-shield or dashboard so the passenger cannot hit the hard objects.c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshieldor dashboard, the air bag gives some increasing the time for the slowing process and thus de-creasing the average force on the passenger.d.The airbag is there to insure the seatbelt holds
Answers
Answer:
c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshield or dashboard, the air bag gives some increasing the time for the slowing process and thus decreasing the average force on the passenger
Explanation:
As soon as the vehicle collides the air bags open up almost instantly within a fraction of seconds.The air bag reduces the force of impact by providing the longer time for the change in momentum in accordance with the Newton's second law and distributes the force over a larger area of impact therefore reducing the pressure.The impulse in each case remains the same because when the air bags are engaged in the process the force is lesser but the time involved is more and we know that impulse is the product of force and time.The stopping impulse is the same for either the hard objects or the airbag.
Unlike the windshield or dashboard, the airbag gives some increasing the
time for the slowing process and thus decreasing the average force on the
passenger.
Air bags are put in automobiles ion order to prevent life threatening injuries
such as brain damage to individuals. The air bag contains air helps
to increase the time for the slowing process.
This thereby prevents the average force on the passenger and prevents
injuries which should have occurred.
Read more on https://brainly.com/question/14781865
In the microscopic view of electrical conduction in a copper wire, electrons are accelerated by an electric field and then collide with metal atoms after traveling about4.2×10−8m.
If an electron begins from rest and is accelerated by a field of 0.080 N/C, what is its speed when it collides with a metal atom?
Answers
Answer:
34.35464 m/s
Explanation:
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
E = Electric field = 0.08 N/C
s = Displacement = [tex]4.2\times 10^{-8}\ m[/tex]
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Electrical force is given by
[tex]F=qE[/tex]
Work done is given by
[tex]W=Fs\\\Rightarrow W=qEs[/tex]
Work done is also given by the kinetic energy
[tex]\dfrac{1}{2}mv^2=qEs\\\Rightarrow v=\sqrt{\dfrac{2qEs}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 0.08\times 4.2\times 10^{-8}}{9.11\times 10^{-31}}}\\\Rightarrow v=34.35464\ m/s[/tex]
The velocity of the electron is 34.35464 m/s
A charged comb often attracts small bits of dry paper that then fly away when they touch the comb. Explain why that occurs.
Answers
Explanation:
Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.
A sinusoidal sound wave moves through a medium and is described by the displacement wave function
s(x, t) = 2.00 cos (15.7x ? 858t)
where s is in micrometers, x is in meters, and t is in seconds. Find (a) the amplitude, (b) the wavelength, and (c) the speed of this wave.
Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x
Answers
Answer
given,
displacement wave function
s(x, t) = 2.00 cos (15.7 x + 858 t)
now, comparing the wave equation with general equation.
s(x, t) = A cos (k x + ω t)
where A is the amplitude of the wave in micrometer.
now,
a) Amplitude of the wave
A = 2 x 10⁻⁶ μ m
b) [tex]\lambda = \dfrac{2\pi}{k}[/tex]
k = 15.7 m
[tex]\lambda = \dfrac{2\pi}{15.7}[/tex]
λ = 0.4 m
c) wave speed
[tex]v = \dfrac{\omega}{k}[/tex]
[tex]v = \dfrac{858}{15.7}[/tex]
v = 54.65 m/s
d) For instantaneous displacement
Assuming the position and time is given as
x = 0.05 m and t = 3 m s
now,
s(x, t) = 2.00 cos (15.7 x 0.05 + 858 x 3 x 10⁻³ )
s(x, t) = 2.00 cos (3.359)
s(x,t) = -1.95 μ m
The amplitude of the wave is 2.00 micrometers, the wavelength is 15.7, and the speed of the wave is approximately 54.65 m/s.
Explanation:The displacement wave function is given as s(x, t) = 2.00 cos (15.7x - 858t). In this wave function, the amplitude is the coefficient in front of the cosine function, which is 2.00 micrometers. The wavelength can be determined by finding the coefficient of the x term inside the cosine function, which is 15.7. The speed of the wave can be calculated by dividing the angular frequency (in this case, 858) by the wave number (in this case, 15.7), resulting in a speed of approximately 54.65 m/s.
Learn more about Wave properties here:https://brainly.com/question/32429112
#SPJ3
A water gun fires 5 squirts per second. The speed of the squirts is 15 m/s.
By how much distance is each consecutive squirt
separated?
Answers
Final answer:
Each consecutive squirt from the water gun is separated by 3 meters, calculated by multiplying the speed of the squirts (15 m/s) by the interval between each squirt (0.2 seconds).
Explanation:
The question is asking to calculate the distance at which each consecutive squirt from a water gun is separated when the water gun fires squirts at a certain rate and speed. Given that the water gun fires 5 squirts per second and the speed of the squirts is 15 m/s, we can use the formula distance = speed × time to find the separation distance between squirts.
Since there are 5 squirts per second, each squirt is separated by 1/5 of a second, or 0.2 seconds. To find the separation distance, we multiply the speed of the squirts by the time interval between each squirt:
Distance = Speed × Time
Distance = 15 m/s × 0.2 s
Distance = 3 meters
Therefore, each consecutive squirt is separated by 3 meters.
A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where tis in s. Its initial position is x0 = 2.3m at t0 = 0 s
A)
At 2.2s , what is the particle's position?
Express your answer with the appropriate units.
B)
At 2.2s , what is the particle's velocity?
Express your answer with the appropriate units.
C)
At 2.2s , what is the particle's acceleration?
Answers
Answer:
A) At 2.2 s the position of the particle is 9.4 m.
B) At t =2.2 s the velocity is 9.7 m/s.
C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²
Explanation:
Hi there!
A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:
dx/dt = 2 · t²
Separating variables, we can find the equation of position as a function of time:
dx = 2 · t² · dt
Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:
∫ dx = 2 ∫ t² · dt
x - 2.3 m = 2/3 · t³
x = 2.3 m + 2/3 m/s³ · t³
Replacing t = 2.2 s:
x = 2.3 m + 2/3 m/s³ · (2.2 s)³
x = 9.4 m
At 2.2 s the position of the particle is 9.4 m
B) Now, let´s evaluate the velocity function at t = 2.2 s:
v = 2 · t²
v = 2 m/s³ · (2.2 s)²
v = 9.7 m/s
At t =2.2 s the velocity is 9.68 m/s
C) The acceleration is the variation of the velocity over time (the derivative of the velocity):
dv/dt = a
a = 4 · t
At t = 2.2 s:
a = 4 m/s³ · 2.2 s
a = 8.8 m/s²
At t = 2.2 s the acceleration of the particle is 8.8 m/s²
(A) The particle's position at time, t = 2.2 s is 7.1 m.
(B) The velocity of the particle at 2.2 s is 9.68 m/s.
(C) The acceleration of the particle at 2.2 s is 8.8 m/s².
The given parameters:
Velocity, Vx = 2t² m/sInitial position of the particle, X₀ = 2.3 mThe particle's position at time, t = 2.2 s is calculated as follows;
[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]
The velocity of the particle at 2.2 s is calculated as follows;
[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]
The acceleration of the particle at 2.2 s is calculated as follows;
[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]
Learn more about velocity and acceleration here: https://brainly.com/question/139187
The work function of palladium is 5.22 eV.
(a) What is the minimum frequency of light required to observe the photoelectric effect on Pd?
(b) If light with a 200 nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons?
Answers
a) The minimum frequency of the light must be [tex]1.26\cdot 10^{15} Hz[/tex]
b) The maximum velocity of the electrons is [tex]5.93\cdot 10^5 m/s[/tex]
Explanation:
a)
The photoelectric effect is a phenomenon that occurs when electromagnetic radiation hits the surface of a metal causing the release of electrons from the metal's surface.
The equation of the photoelectric effect is:
[tex]hf = \phi +K_{max}[/tex]
where :
[tex]hf[/tex] is the energy of the incoming photons, where
[tex]h[/tex] is the Planck's constant
[tex]f[/tex] is the frequency of the incoming photons
[tex]\phi[/tex] is the work function of the metal, the minimum energy that the photons must have in order to be able to free electrons from the metal
[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted electrons
In order to free electrons, the minimum energy of the photons must be at least equal to the work function (so that the kinetic energy of the electrons is zero, [tex]K_{max}=0[/tex]. Therefore,
[tex]h f_0 = \phi[/tex]
In this case,
[tex]\phi = 5.22 eV \cdot (1.6\cdot 10^{-19})=8.35\cdot 10^{-19} J[/tex]
Therefore, the minimum frequency of the photons must be
[tex]f_0 = \frac{\phi}{h}=\frac{8.35\cdot 10^{-19}}{6.63\cdot 10^{-34}}=1.26\cdot 10^{15} Hz[/tex]
b)
In this case, the wavelength of the incoming light is
[tex]\lambda = 200 nm = 200 \cdot 10^{-9} m[/tex]
We can find the frequency by using the wave equation:
[tex]f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{200\cdot 10^{-9}}=1.5\cdot 10^{15} Hz[/tex]
Now we can use the equation of the photoelectric effect to find the maximum kinetic energy of the electrons:
[tex]K_{max} = hf-\phi = (6.63\cdot 10^{-34})(1.5\cdot 10^{15})-8.35\cdot 10^{-19}=1.60\cdot 10^{-19} J[/tex]
And therefore, we can find their velocity by using the equation for the kinetic energy:
[tex]K_{max} = \frac{1}{2}mv^2[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electrons
v is their speed
Solving for v,
[tex]v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})}{9.11\cdot 10^{-31}}}=5.93\cdot 10^5 m/s[/tex]
Learn more about photoelectric effect:
brainly.com/question/10015690
#LearnwithBrainly
A railroad freight car, mass 15 000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50 000-kg loaded second car, initially at rest and with brakes released. What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision
Answers
Answer:
23.0760769 %
Explanation:
[tex]m_1[/tex] = Mass of freight car = 15000 kg
[tex]m_2[/tex] = Mass of second car = 50000 kg
[tex]v_1[/tex] = Velocity of freight car = 2 m/s
[tex]v_2[/tex] = Velocity of second car = 0
v = Combined mass velocity
As the linear momentum of the system is conserved we have
[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]
The initial kinetic energy
[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]
Final kinetic energy
[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]
The percentage is given by
[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]
The change in percentage of initial kinetic energy is 23.0760769 %
By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 3.0 N ?
Answers
To solve this problem we will apply the definition of electrostatic force. From the variables present there and explained later we will find the value of the distance reorganizing said expression, that is
[tex]F = \frac{k q_1 q_2}{d^2}[/tex]
Here
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object
d = Distance
Replacing our values we have that
[tex]3 = \frac{(9*10^9)(1)(1)}{d^2}[/tex]
Rearranging and solving for the distance we have
[tex]d = 54772.25m[/tex]
Therefore the distance between the two objects is 54772.25m
If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux that planet B receives.
Answers
Answer:
The nearest plant (A) receives 4 times more radiation from the farthest plant
Explanation:
The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of the sphere
I = P / A
P = I A
The area of the sphere is
A = 4π r²
Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets
P = I₁ A₁ = I₂ A₂
I₁ / I₂ = A₂ / A₁
Suppose index 1 corresponds to the nearest planet,
r2 = 2 r₁
I₁ / I₂ = r₁² / r₂²
I₁ / I₂ = r₁² / (2r₁)²
I₁ / I₂ = ¼
4 I₁ = I₂
The nearest plant (A) receives 4 times more radiation from the farthest plant
An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. Suppose that a typical car rounds the curve with a speed of 11.7m/s and that the radius of the curve is 50.0m. At what angle should the curve be banked?
Answers
To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,
[tex]T_x = nsinA = \frac{mv^2}{r}[/tex]
[tex]T_y = ncosA = mg[/tex]
Dividing both.
[tex]tan A = \frac{v^2}{rg}[/tex]
[tex]tan A = \frac{11.7^2}{50*9.8}[/tex]
[tex]A = tan^{-1} (0.279367)[/tex]
[tex]A = 15.608\°[/tex]
Therefore the angle that should the curve be banked is 15.608°
Vector vector A has a magnitude A and is directed at an angle theta measured with respect to the positive x-axis. What is the magnitude of vector A sub x, the x-component of vector A?
Answers
Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]
Explanation:
A vector in this situation have two components
1) The X-component
2) The Y-component
So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be
[tex]A_x[/tex] = Acos[tex]\theta[/tex]
I hope this will answer your question,
An image is also provided please have a look at that.
Thank you.
Answer: Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
Explanation:
The attached image is a pictorial representation of the question:
From the attached image,
Cos(theta) = adjacent/hypothenus
Cos(theta) = Ax/A
Making Ax the subject of formula,
Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
The center of mass of a pitched baseball or
radius 3.91 cm moves at 33.6 m/s. The ball
spins about an axis through its center of mass
with an angular speed of 52.1 rad/s.
Calculate the ratio of the rotational energy
to the translational kinetic energy. Treat the
ball as a uniform sphere.
Answers
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]
Learn more about kinetic energy:
brainly.com/question/6536722
#LearnwithBrainly
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]
learn more about mass here: https://brainly.com/question/16824076
Two charges are located in the x – y plane. If q 1 = − 3.65 nC and is located at ( x = 0.00 m , y = 0.600 m ) , and the second charge has magnitude of q2 = 4.20 nC and is located at ( x = 1.10 m , y = 0.800 m ) , calculate the x and y components, Ex and Ey , of the electric field, → E , in component form at the origin, ( 0 , 0 ) . The Coulomb force constant is 1/(4πϵ0 ) = 8.99 × 10^9 N ⋅ m^2 /C^2.
Answers
Answer:
Ex = -16.51 N/C Ey = 79.14 N/C
Explanation:
As the electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.
So, we can first the field due to q1.
Due to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward,(like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows.
E₁ = k*(3.65 nC) / r₁²
If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:
E₁ₓ = 0 E₁y = 8.99*10⁹*3.65*10⁻⁹ / (0.600)²m² = 91.15 N/C (1)
For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.
It will be just the pythagorean distance between the points located at the coordinates (1.10 m, 0.800 m) and (0,0), as follows:
r₂² = 1.10²m² + (0.800)²m² = 1.85 m²
The magnitude of the electric field due to q2 can be found as follows:
E₂ = k*q₂ / r₂² = 8.99*10⁹*(4.2)*10⁹ / 1.85 = 20.41 N/C (2)
Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.
In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:
E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ
the cosine of θ, is just, by definition, the opposite of x/r₂:
⇒ cos θ =- (1.10 m / √1.85 m²) =- (1.10 / 1.36) = -0.809
By the same token, sin θ can be obtained as follows:
sin θ = - (0.800 m / 1.36 m) = -0.588
⇒E₂ₓ = 20.41 N/C * (-0.809) = -16.51 N/C (pointing to the left) (3)
⇒E₂y = 20.41 N/C * (-0.588) = -12.01 N/C (pointing downward) (4)
The total x and y components due to both charges are just the sum of the components of Ex and Ey:
Ex = E₁ₓ + E₂ₓ = 0 + (-16.51 N/C) = -16.51 N/C
From (1) and (4), we can get Ey:
Ey = E₁y + E₂y = 91.15 N/C + (-12.01 N/C) =79.14 N/C
Answer:
Ex = 15.505 N/C
Ey = 79.144 N/C
Explanation:
Particle 1
[tex]E_{1} = \frac{k*q_{1} }{r^2_{1}} \\\\r^2_{1} = 1.1^2+0.8^2\\\\r^2_{1} = 1.85 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.8}{1.1}) = 36.03 degree\\\\ E_{1} = \frac{(8.99*10^9)*(4.2*10^(-9)) }{1.85}\\\\E_{1} = 20.4096 N/C[/tex]
Away from the particle at (0,0) due to + charge
Particle 2
[tex]E_{2} = \frac{k*q_{2} }{r^2_{2}} \\\\r^2_{2} = 0.36 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.6}{0}) = 90 degree\\\\ E_{2} = \frac{(8.99*10^9)*(3.65*10^(-9)) }{0.36}\\\\E_{2} = 91.149 N/C[/tex]
Towards from the particle at (0,0) due to - charge
Resultant Electric field in y direction
[tex]E_{res,y} = E_{2} -E_{1}*cos(Q)\\E_{res,y} = (91.149) - (20.4096)*sin(36.03)\\\\E_{res,y} = 79.144 N/C[/tex]
Resultant Electric field in x direction
[tex]E_{res,x} = E_{1}*cos(Q)\\E_{res,y} =(20.4096)*cos(36.03)\\\\E_{res,x} = 16.505 N/C[/tex]
A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the velocity of the skater?
a. -0.07 m/s
b. 0.15 m/s
c. 0.30 m/s
d. 0.07 m/s
Answers
Answer:
(d) 0.07 m/s
Explanation:
Given Data
Snowball mass m₁=0.15 kg
Ice skater mass m₂=65.0 kg
Snowball velocity v₁=32.0 m/s
To find
Velocity of Skater v₂=?
Solution
From law of conservation of momentum
[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]
So Option d is correct one
The velocity of skater is [tex]0.07m/s[/tex]
Option d is correct.
Law of conservation of momentum:For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.
The expression is given as,[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
Given that, mass of ice skater [tex]m_{1}=65kg[/tex], mass of snow ball [tex]m_{2}=0.15kg[/tex] Velocity of snow ball [tex]v_{2}=32m/s[/tex]Substitute values in above expression.[tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]
The velocity of skater is [tex]0.07m/s[/tex]
Learn more about the Velocity here:
https://brainly.com/question/24445340
2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.
Answers
The specific kinetic energy of a mass (in kilojoules per kilogram) moving at a velocity of 40 m/s is calculated using the kinetic energy formula, resulting in 0.8 kJ/kg.
Explanation:The specific kinetic energy of a mass moving with a velocity can be determined using the formula for kinetic energy, K = 0.5*m*v². In this case, where the velocity 'v' is given as 40 m/s, and we want to solve the kinetic energy per kilogram, we can consider the mass 'm' as 1 kg. Hence the specific kinetic energy would be K = 0.5*(1 kg)*(40 m/s)² = 800 J = 0.8 kJ/kg, because 1 kilojoule (kJ) = 1000 joules (J).
Learn more about Specific Kinetic Energy here:https://brainly.com/question/30832620
#SPJ3
After a parallel-plate capacitor has been fully charged by a battery, the battery is disconnected and the plate separation is increased. Which of the following statements is correct? Please explain in detail why the staement is correct!
A) The energy stored in the capacitor increases.
B) The charge on the plates increases.
C) The charge on the plates decreases.
D) The potential difference between the plated decreases.
E) The energy stored in the capacitor decreases.
Answers
Answer:
A) The energy stored in the capacitor increases.
Explanation:
For a capacitor fully charged by battery, when disconnected from battery and the plate separation is increased. The charge on the plate remain constant because there is no where for it to go( it has been disconnected from battery). But the capacitance would decrease, while also the potential difference would increase.
Q = CV ....1
Q is the charge, C is capacitance, V is the potential difference.
The energy stored in a capacitor is given by:
E = 1/2 CV^2
E = 1/2 QV .......2
E is the energy stored in the capacitor,
Therefore since Q remain constant and V increases when the distance between the plates is increased, then according to the equation 2 above the energy stored in the capacitor increases.
What is the correct answer
Answers
Answer:
D and compound
Explanation:
because N2 is = to a compound
A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude 5.0 rad/s2. How many revolutions does it make while slowing down before coming to rest
Answers
Answer:
It takes 6.37 revolutions to stop.
Explanation:
The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0[tex] \frac{rad}{s^{2}}[/tex].Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:
[tex]\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta [/tex]
with [tex] \omega[/tex] the final angular velocity (is zero because the wheel comes to rest), [tex] \omega_{i}[/tex] the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :
[tex]\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha} [/tex]
[tex]\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad [/tex]
The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:
[tex]40rad=\frac{40rad}{2\pi\frac{rad}{rev}} [/tex]
[tex] 40rad=6.37 rev[/tex]
Number of revolution take place is 6.4 revolution
Given that;Velocity of rotating wheel = 20 rad/s
Acceleration of magnitude = 5.0 rad/s²
Find:Number of revolution take place
Computation:Using Third equation of motion;
v² - u² = 2as20² - 0² = 2(5)(s)
400 = (10)(s)
Total distance = 40 rad
Number of revolution take place = Total distance / 2π
Number of revolution take place = 40 / 2(3.14)
Number of revolution take place = 40 / 6.28
Number of revolution take place = 6.369
Number of revolution take place = 6.4 revolution
Learn more:
https://brainly.com/question/17773408?referrer=searchResults
If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?
Answers
Yes, it is possible. The ball would fall straight down from the perspective of a person standing at the side of the road. The person who threw the ball would see it initially move backward and then fall vertically downward.
Explanation:Yes, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road while someone is riding in the back of a pickup truck and throws a softball straight backward. This would occur under the condition that the ball is thrown with the same initial velocity as the truck's speed. When the ball is thrown straight backward with the same initial speed as the truck's, it will continue to move with the same speed in the backward direction relative to the truck. From the perspective of the person who threw the ball, they would see the ball initially move straight backward and then fall vertically downward due to the force of gravity.
Learn more about Motion of a ball thrown backward from a moving vehicle here:https://brainly.com/question/36599622
#SPJ12
The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.
To the person who threw the ball, it would appear to travel straight backward relative to them.
If someone is riding in the back of a pickup truck and throws a softball straight backward, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road. This will occur if the velocity at which the ball is thrown backward is equal to the velocity of the truck moving forward. In this case, the forward motion of the truck and the backward motion of the ball will cancel each other out.
For the person who threw the ball, the motion would still appear as if the ball was thrown straight backward relative to their frame of reference, assuming the truck is moving with constant velocity. The ball would move backward with the same speed at which it was thrown.
To summarize:
The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.
To the person who threw the ball, it would appear to travel straight backward relative to them.
The magnitude of the electric force between two protons is 2.30 x 10^-26 N. How far apart are they?
a) 0.0220 mb) 0.100 mc) 0.480 md) 0.000570 me) 3.10 m
Answers
To solve this problem we will use the concepts given by Coulomb's law defined for force, said law is mathematically described as
[tex]F = \frac{kq_1 q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex]= Charge of two protons
r = Distance between them
F = Force
Our values are given as,
[tex]F =2.3*10^{-26} N[/tex]
[tex]q_1 =q_2 = 1.6*10^{-19} C[/tex]
[tex]k =9*10^9 Nm^2/C2[/tex]
Rearrenging to find the distance and replacing we have that
[tex]r^2=\frac{(9*10^9 )(1.6*10^{-19})^2 }{2.3*10^{-26} }[/tex]
[tex]r^2=10.01*10^{-3} m^2[/tex]
[tex]r =\sqrt{10.01*10^{-3} m^2}[/tex]
[tex]r = 0.100m[/tex]
Therefore the correct option is B.
A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a long, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Answers
Answer:
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
Explanation:
If we define the magnitude of the field as B, then we have:
Total magnitude of the field [tex]B_{t}[/tex] = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.
B_loop = (μ0)*([tex]I_{2}[/tex])/2*R
B_wire = (μ0)*([tex]I_{1}[/tex])/2*π*D
Thus:
B_loop = B_wire at the center of the loop.
(μ0)*([tex]I_{2}[/tex])/2*R = (μ0)*([tex]I_{1}[/tex])/2*π*D
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
The magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
The given parameters;
radius of the loop = Rcurrent in the loop, I = I₂distance of the loop from the wire, = DThe magnetic field at the center of the loop is calculated as follows;
[tex]B_o = \frac{\mu_o I_2}{2R}[/tex]
The magnetic field at the distance of the wire is calculated as follows;
[tex]B_o = \frac{\mu_o I_1}{2\pi D}[/tex]
The magnitude of current at the center of the loop is calculated as follows;
[tex]\frac{\mu_o I_2 }{2R} = \frac{\mu_o I_1}{2\pi D} \\\\I_1 = \frac{\pi D I_2}{R}[/tex]
Thus, the magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
Learn more here:https://brainly.com/question/12904923
An equilateral triangle with side lengths of 0.50 m has a 5.0 nC charge placed at each corner. What is the magnitude of the electric field at the midpoint of one of the three sides? (A) 240 N/C (B) 180 N/C (C) 720 N/C (D) 480 N/C (E) 120 N/C
Answers
To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,
[tex]x = \frac{\sqrt{3}}{2}a[/tex]
[tex]x = \frac{\sqrt{3}}{2}(0.5)[/tex]
[tex]x = 0.43301m[/tex]
The magnitude of the electric field is
E=\frac{kQ}{x^2}
Here,
k = Coulomb's Constant
Q = Charge
x = Distance
[tex]E=\frac{(9*10^9)(5*10^{-9})}{(0.43301)^2}[/tex]
[tex]E = 240.002N/C \approx 240N/C[/tex]
Therefore the correct answer is A.
During an ice show a 60 kg skater leaps into the air and is caught by an initially stationary 75.0 kg skater.a. What is their final velocity assuming negligible friction and that the 60.0-kg skater's original horizontal velocity is 4.00 m/s?b. How much kinetic energy is lost?
Answers
Answer:
(a). The final velocity is 1.78 m/s.
(b). The lost kinetic energy is 266.13 J.
Explanation:
Given that,
Mass of first skater = 60 kg
Mass of second skater = 75.0 kg
Initial velocity = 4.00 m/s
(a). We need to calculate the final velocity
Using conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]
Put the value into the formula
[tex]60\times4.00+75.0\times0=(60+75.0)v[/tex]
[tex]v=\dfrac{60\times4.00}{60+75}[/tex]
[tex]v=1.78\ m/s[/tex]
The final velocity is 1.78 m/s.
(b). We need to calculate the lost kinetic energy
Using formula of kinetic energy
[tex]\Delta E=E_{2}-E_{1}[/tex]
[tex]\Delta E=\dfrac{1}{2}(m_{1}u_{1}^2+m_{2}u_{2}^2)-\dfrac{1}{2}(m_{1}+m_{2})v^{2}[/tex]
[tex]\Delta E=\dfrac{1}{2}(60\times4^2)-\dfrac{1}{2}\times(60+75)\times1.78^2[/tex]
[tex]\Delta E=266.13\ J[/tex]
Hence, (a). The final velocity is 1.78 m/s.
(b). The lost kinetic energy is 266.13 J.
a. The final velocity should be considered as the 1.78 m/s.
b. The lost kinetic energy should be considered as the 266.13 J.
Conservation of momentum:Since
Mass of first skater = 60 kg
Mass of second skater = 75.0 kg
Initial velocity = 4.00 m/s
a.
We know that
[tex]mm1v1 + m2v2 = (m1 + v1)v\\\\60*4.00 + 75*0 = (60 + 75)v[/tex]
v = 60 + 4/60 + 75
= 1.78 m/s
b.
[tex]E = 1/2(m1v1^2 + m2v2^2) - 1/2(m1 + vm2)v^2\\\\= 1/2(60*4^2) - 1/2 * (60 + 75)*1.78^2[/tex]
= 266.13 J
learn more about kinetic energy here: https://brainly.com/question/16824076
According to the U.S. Green Building Council, what percentage of the world’s energy use and greenhouse gas emissions can be attributed to buildings?
Answers
According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions
Answer:
40%
Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.