Answer:
[tex]\frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]
Explanation:
The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere
[tex]V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3[/tex]
[tex]V = \frac{4}{3}\pi(r_2^3 - r_1^3)[/tex]
The mass of the shell is the product of the volume and its density
[tex]m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 39 m/s m/s . The plane is in a 17 m/s m/s wind blowing toward the southwest relative to the earth. For help with math skills, you may want to review: Vector Addition Resolving Vector Components For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Flying in a crosswind. Part A Letting x be east and y be north, find the components of v ⃗ P/E v→P/E (the velocity of the plane relative to the earth). Express your answers in meters per second separated by a comma. View Available Hint(s) v x vx , v y vy = nothing m/s m/s Submit Part B Find the magnitude of v ⃗ P/E v→P/E . Express your answer in meters per second. View Available Hint(s) v P/E vP/E = nothing m/s m/s Submit Part C Find the direction of v ⃗ P/E v→P/E . Express your answer in degrees. View Available Hint(s) ϕ ϕ = nothing ∘ ∘ south of west Submit Provide Feedback Next
Answer:
A. [tex]|\vec v_t|=52.42\ m/s[/tex]
B. [tex]\theta=256.74^o[/tex]
Explanation:
Velocity Vector
The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.
We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as
[tex]v_{xp}=|v_p|cos\alpha_p[/tex]
[tex]v_{yp}=|v_p|sin\alpha_p[/tex]
[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]
[tex]v_{xp}=39cos\ 270^o=0[/tex]
[tex]v_{yp}=39sin\ 270^o=-39[/tex]
Thus, the velocity of the plane is
[tex]\vec v_p=0\hat i-39\hat j[/tex]
The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):
[tex]v_{xw}=|v_w|cos\alpha_w[/tex]
[tex]v_{yw}=|v_w|sin\alpha_w[/tex]
[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]
[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]
Thus, the velocity of the wind is
[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]
Now we perform the vector addition to compute the plane's final speed
[tex]\vec v_t=\vec v_p+\vec v_w[/tex]
[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]
[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]
A) The magnitude of the total velocity is
[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]
[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]
B) The direction angle is given by
[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]
[tex]\theta=arctan\ 4.24[/tex]
[tex]\theta=76.74^o[/tex]
This angle is west of south, we must add 180° to express it in due reference, thus
[tex]\boxed{\theta=256.74^o}[/tex]
If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100 mm length of the axon?
Answer:
Charge enter a 0.100 mm length of the axon is [tex]8.98\times 10^{-12} C[/tex]
Explanation:
Electric field E at a point due to a point charge is given by
[tex]E=k \frac{q}{r^2}[/tex]
where [tex]k[/tex] is the constant =[tex]9.0 \times 10^9 Nm^2 / C^2[/tex]
[tex]q[/tex] is the magnitude of point charge and [tex]r[/tex] is the distance from the point charge
Charges entering one meter of axon is [tex]5.\times 10^{11} \times (+e)[/tex]
Charges entering 0.100 mm of axon is [tex]5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}[/tex]
substituting the value of [tex]+e=1.6\times 10^{-19} C[/tex] in above equation, we get charge enter a 0.100 mm length of the axon is
[tex]q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C[/tex]
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horizontal direction, determine the Cartesian representation of v→ relative to the (iˆ, jˆ) component system.
Answer:
[tex]\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
Explanation:
The x- and y- components of the velocity vector can be written as following:
[tex]\vec{v}_x = ||\vec{v}||\cos(\theta)\^i[/tex]
[tex]\vec{v}_y = ||\vec{v}||\sin(\theta)\^j[/tex]
Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:
[tex]\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
Tim and Rick both can run at speed v_r and walk at speed v_w, with v_r > v_w They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.
How long does it take Rick to cover the distance D?
Express the time taken by Rick in terms of v_r, v_w, and D.
Find Rick's average speed for covering the distance D.
Express Rick's average speed in terms of v_r and v_w.
How long does it take Tim to cover the distance?
Express the time taken by Tim in terms of v_r, v_w, and D.
Who covers the distance D more quickly?
In terms of given quantities, by what amount of time, Delta t, does Tim beat Rick?
It will help you check your answer if you simplify it algebraically and check the special case v_r = v_w
Express the difference in time, Delta t in terms of v_r, v_w, and D.
In the special case that v_r = v_w, what would be Tim's margin of victory Delta t(v_r = v_w)?
What is the electric field strength just outside the surface of a conducting sphere carrying surface charge density 1.4 μC/m2μC/m2?
Answer:
[tex]E=158.19\frac{kN}{m}[/tex]
Explanation:
Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]
The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]
Recall that the surface charge density is defined as:
[tex]\sigma=\frac{q}{S}[/tex]
Now, we get the electric field strength:
[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]
The electric field strength just outside the surface of a conducting sphere with surface charge density of 1.4 μC/m^2 can be calculated using Gauss' Law. This gives an electric field strength of approximately 1.58 x 10^5 N/C.
Explanation:The electric field strength just outside the surface of a conducting sphere, with a surface charge density of 1.4 μC/m2, can be calculated using Gauss' Law, which states that the electric field is equal to the surface charge density divided by the permittivity of free space, ε0.
It is given by the formula:
E = σ/ε0
Where E is the electric field strength, σ is the charge density, and ε0 is the permittivity of free space. Using the given value for σ (1.4 x 10-6 C/m2) and the known value for ε0 (8.85 x 10-12 C2/N.m2), we find that:
E = (1.4 x 10-6) / (8.85 x 10-12)
This simplifies to approximately E = 1.58 x 105 N/C.
Therefore, the electric field strength just outside the surface of the conducting sphere is approximately 1.58 x 105 N/C.
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A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 21.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.10 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 30.0 m before hitting the water.
After leaving the edge of the cliff how much time does the diver take to get to the water?
How far horizontally does the diver travel from the cliff face before hitting the water?
Remember that the angle is at a downward slope to the right.
Final answer:
The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.
Explanation:
To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.
To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.
An object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. The object then accelerates at a rate of -3 rad/s2 until it stops. Through what angular displacement (in rad) does the object move between when it starts moving and when it stops?
Answer:
Total angular displacement will be 240 radian
Explanation:
In first case object starts from rest so initial angular speed [tex]\omega _i=0rad/sec[/tex]
Angular acceleration is given [tex]\alpha =2rad/sec^2[/tex]
Final angular speed[tex]\omega _f=24rad/sec[/tex]
From third equation of motion [tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]
So [tex]24^2=0^2+2\times 2\times \Theta[/tex]
[tex]\Theta =144[/tex] radian
Now in second case as the objects finally stops
So final velocity [tex]\omega _f=0rad/sec[/tex]
Angular acceleration [tex]\alpha =-3rad/sec^2[/tex]
So [tex]0^2=24^2-2\times 3\times \Theta[/tex]
[tex]\Theta =96[/tex] radian
So total angular displacement will be 96+144 = 240 radian
To find the angular displacement, we can analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. We can calculate the time and angular displacement in each phase using the formulas for angular speed and angular displacement. Finally, we can add the angular displacements from both phases to find the total angular displacement.
Explanation:To find the angular displacement, we need to analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. In the first phase, the object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. We can use the formula:
Final Angular Speed = Initial Angular Speed + Angular Acceleration * Time
Using this formula, we can find the time taken in the first phase. Then, we can calculate the angular displacement during the first phase using the formula:
Angular Displacement = Initial Angular Speed * Time + 0.5 * Angular Acceleration * Time^2
In the second phase, the object decelerates at a rate of -3 rad/s^2 until it stops. We can use the same formulas to find the time and angular displacement in the second phase. Finally, we can add the angular displacements from both phases to get the total angular displacement.
The total angular displacement will be the sum of the angular displacements in the two phases.
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An object whose weight is 100lbf( pound force) experiences a decrease i kinetic energy of 500ft-lb, and an increase in potential energy of 1500ft-lbf. The initial velocity and elevation of the object, each relative to the surface of the Earth are 40ft/s and 30ft.
(a). Find final velocity in ft/s . (b) find final elevation.
The final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].
The potential energy is the energy possessed by virtue of configuration and the kinetic energy is the energy possessed due to motion.
Given:
Initial velocity, [tex]u=40 ft/s[/tex]
Initial elevation, [tex]h_i=30 ft[/tex]
Weight of the object, [tex]w=100lbf[/tex]
Increase in potential energy, [tex]\Delta PE=1500 ft{-}lbf[/tex]
Decrease in kinetic energy, [tex]\Delta KE=-500 ft{-}lbf[/tex]
(a)
The final velocity of the object is computed as:
[tex]\frac{1}{2} \frac{w}{g}v^2-\frac{1}{2} \frac{w}{g}u^2= \Delta KE\\v^2=40^2 + 2 \times \frac{32.17}{100} \times (-500)\\v= \sqrt{1278.3}\\v=35.75 \ ft/s[/tex]
(b)
The final elevation of the object is computed as:
[tex]mgh_f-mgh_i= \Delta PE\\h_f= h_i+ \frac {\Delta PE}{mg}\\h_f= 30 + \frac{1500}{100}\\h_f=45 ft[/tex]
Therefore, the final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].
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A gun is fired with angle of elevation . What is the muzzle speed if the maximum height of the shell is 500 m?
Answer:
The muzzle speed u =200 m/s
Explanation:
Let u be the muzzle speed.
maximum height s= 500 m
Assuming angle of elevation to be 30°
then initial vertical speed = usin30° = u/2
moreover, when the bullet reaches the maximum height its vertical velocity will be zero.
then using
[tex]v^2=u^2-2as[/tex]
a= 10 m/s^2
u= u/2
v= 0
s=500
we get
[tex]0^2=u^2/4-2\times10\times500[/tex]
u= 200 m/s.
Therefore, muzzle speed = 200 m/s
Answer:
The muzzle speed is 197.9 m/s.
Explanation:
Given that,
Maximum height = 500 m
Suppose the angle is 30°.
We need to calculate the muzzle speed
Using formula of maximum height
[tex]h_{max}=\dfrac{u^2}{2g}[/tex]
Where, h = maximum height
g = acceleration due to gravity
u = initial vertical velocity
Put the value into the formula
[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]
[tex]u^2=8\times500\times9.8[/tex]
[tex]u=\sqrt{4\times500\times9.8}[/tex]
[tex]u=197.9\ m/s[/tex]
Hence, The muzzle speed is 197.9 m/s.
The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck’s rear bumper. The car accelerates at a constant 0.600 m/s2, then pulls back into the truck’s lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5 m long, and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?
Answer:
a) 15.864s
b) 392.78m
c) 29.52 m/s
Explanation:
The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is
distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)
= 24 + 21 + 26 + 4.5 = 75.5 m
As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck
[tex]s = at^2/2[/tex]
[tex]75.5 = 0.6t^2/2[/tex]
[tex]t^2 = 251.67[/tex]
[tex]t = \sqrt{251.67} = 15.864s[/tex]
b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time
= 75.5 + 20*15.864 = 392.78 m
c) final speed of the car is the initial speed plus the change in speed
[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]
To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.
Explanation:u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.
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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground.
A. How far does the shot travel?
B. Repeat the calculation of the first part for angle 42.5? .
C. Repeat the calculation of the first part for angle 45 ? .
D. Repeat the calculation of the first part for angle 47.5? .
E. At what angle of release does she throw the farthest?
A) Horizontal range: 16.34 m
B) Horizontal range: 16.38 m
C) Horizontal range: 16.34 m
D) Horizontal range: 16.07 m
E) The angle that gives the maximum range is [tex]41.9^{\circ}[/tex]
Explanation:
A)
The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:
[tex]s=u_y t + \frac{1}{2}at^2[/tex] (1)
where
s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity, where
u = 12.0 m/s is the initial speed
[tex]\theta=40.0^{\circ}[/tex] is the angle of projection
So
[tex]u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s[/tex]
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (downward)
Substituting the numbers, we get
[tex]-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0[/tex]
which has two solutions:
t = -0.21 s (negative, we ignore it)
t = 1.778 s (this is the time of flight)
The horizontal motion is instead uniform, so the horizontal range is given by
[tex]d=u_x t[/tex]
where
[tex]u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s[/tex] is the horizontal velocity
t = 1.778 s is the time of flight
Solving, we find
[tex]d=(9.19)(1.778)=16.34 m[/tex]
B)
In this second case,
[tex]\theta=42.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.1t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.851 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.85)(1.851)=16.38 m[/tex]
C)
In this third case,
[tex]\theta=45^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.5t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.925 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.49)(1.925)=16.34[/tex] m
D)
In this 4th case,
[tex]\theta=47.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.8t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.981 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.11)(1.981)=16.07 m[/tex]
E)
From the previous parts, we see that the maximum range is obtained when the angle of releases is [tex]\theta=42.5^{\circ}[/tex].
The actual angle of release which corresponds to the maximum range can be obtained as follows:
The equation for the vertical motion can be rewritten as
[tex]s-u sin \theta t + \frac{1}{2}gt^2=0[/tex]
The solutions of this quadratic equation are
[tex]t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}[/tex]
This is the time of flight: so, the horizontal range is
[tex]d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta[/tex]
It can be found that the maximum of this function is obtained when the angle is
[tex]\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})[/tex]
Therefore in this problem, the angle which leads to the maximum range is
[tex]\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}[/tex]
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By what order of magnitude is something that runs in nanoseconds faster than something that runs in milliseconds?
To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.
A nanosecond is one billionth of a second while and a millisecond is one millionth of a second
[tex]\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6[/tex]
Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds
Word magnitude means the extent of something. It is the property that determines the object is larger or smaller than the other. The object's magnitude can be arranged into class.
Nano second is the SI unit of time that is equal to bn of a second. That is 10⁻⁹ seconds. The nanosecond is a one billion th of the second, whereas an millisecond is 1000th of a second. The nanosecond process is thus 1,000,000 times faster.The correct answer is 100,00,00.
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Astronomers have not yet reported any Earth-like planets orbiting other stars because (a) there are none; (b) they are not detectable with current technology; (c) no nearby stars are of the type expected to have Earth-like planets; (d) the government is preventing us from reporting their discovery.
Answer:
(b) they are not detectable with current technology
Explanation:
The improvement in exoplanet detection methods in recent years, thanks to spaces telescopes such as the Kepler, has led to a revolution in the field of astronomy. The findings have gone from focusing on Jupiters hot to super-earth and, ultimately, to Earth-like planets.
An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole measured counterclockwise from the electric field direction for which the torque on the dipole is zero. B. Which part of orientation in part (a) is stable and which is unstable?
A. To find the orientation angles of the dipole for which the torque is zero, we need to consider the torque equation for an electric dipole in an external electric field.
The torque ([tex]\(\tau\)[/tex]) acting on an electric dipole (p) in an electric field (E) is given by:
[tex]\[ \tau = p \cdot E \cdot \sin(\theta) \][/tex]
Where:
p is the magnitude of the electric dipole moment,
E is the magnitude of the electric field,
[tex]\(\theta\)[/tex] is the angle between the dipole moment (p) and the electric field (E).
For the torque to be zero, [tex]\(\sin(\theta)\)[/tex] must be zero. This happens when [tex]\(\theta = 0\)[/tex] or [tex]\(\theta = \pi\) (180 degrees)[/tex], as [tex]\(\sin(0) = 0\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex].
So, the two possible orientation angles for which the torque is zero are:
[tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field)
[tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field)
B. Now, let's analyze the stability of these orientations:
1. [tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field):
In this orientation, the dipole moment is aligned with the electric field.
Any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is stable.
2. [tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field):
In this orientation, the dipole moment is opposite to the electric field. Similar to the previous case, any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is also stable.
Both orientations are stable because they represent energy minima. Any deviation from these orientations results in a restoring torque that tries to bring the dipole back to its stable position.
Thus, both orientations ([tex]\(\theta = 0\) degrees and \(\theta = 180\)[/tex] degrees) are stable orientations for the electric dipole in a uniform electric field.
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Final answer:
The torque on an electric dipole in a uniform electric field is zero when the dipole is parallel to the field. This orientation is stable.
Explanation:
In general, the torque vector on an electric dipole, p, from an electric field, E, is given by the equation T = p x E, where the cross product represents the direction of the torque. To find all the orientation angles of the dipole for which the torque is zero, we set the cross product equal to zero:
p x E = 0
The torque is zero when the dipole and electric field vectors are parallel. Therefore, the dipole will experience a torque that tends to align it with the electric field vector. This means that the dipole is in a stable equilibrium when it is parallel to the electric field.
Therefore, the orientation angles that result in a zero torque are all angles that make the dipole parallel to the electric field.
Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Write down the value for the speed of light in metric units: c = _________.
Answer:
6.71 × 10^8 mi/hr
Explanation:
Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.
To write it in the metric system, it has to be converted into miles/hour.
We know that,
1 minute = 60 seconds
60 minutes = 1 hour
1 kilometer = 1000 meter
1 miles = 1.6 kilometer
Now,
= [tex]\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}[/tex]
= 1.08 × 10^12 m/ hr (meter/hour)
= [tex]\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}[/tex]
= 6.71 × 10^8 mi/hr (miles/hour)
Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.
Final answer:
The speed of light in a vacuum is precisely known and has a fixed value of c = 2.99792458 × 10^8 m/s, roughly equal to 3.00 × 10^8 m/s. It's a fundamental constant in physics but is slower in materials, as defined by their index of refraction.
Explanation:
The speed of light in a vacuum is represented by the symbol c and has a fixed value of c = 2.99792458 × 108 m/s, which is essentially 3.00 × 108 m/s when rounded to three significant digits. This constant speed is a fundamental physical quantity and is crucial in many areas of physics including relativity and electromagnetism. It's important to note that the speed of light reduces when it passes through matter, which is characterized by the index of refraction n of the material.
Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff f
Complete question:
Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?
Answer:
Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV
Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz
Explanation:
The kinetic energy (KE) of the emitted photon:
KE = 0.5mv²
m is mass of electron = 9.1 X 10⁻³¹ kg
KE = 0.5 * 9.1 X 10⁻³¹ * (460000)² = 9.628 X 10⁻²⁰ J
in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV
The photon energy of the incoming radiation:
E = hf = hc/λ
c is speed of light (photon) = 3 x 10⁸
h is Planck's constant = 6.626 × 10⁻³⁴ J.s
E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)
E = 31.805 X 10⁻²⁰ J
in eV = 31.805 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.985 eV
Part (a) the work function of the surface
KE = hf - W
where;
W is work function
W = hf - KE
W = 31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J
in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV
Part(b) the cutoff frequency for this surface
W =hf
f = W/h
f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)
f = 3.347 x 10¹⁴ Hz
The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.
Explanation:The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.
The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.
The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected
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At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.9 × 10−15 m. What is the magnitude of the repulsive force pushing these two spheres apart? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
Answer:
Force = 3481.1 N.
Explanation:
Below is an attachment containing the solution.
A trapezoidal channel with 6.0 ft bed width, 3 ft water depth and 1:1 side slope, carries a discharge of 250 ft3/s. Determine whether the flow is supercritical or subcritical.
Answer
given,
width of trapezoidal channel, b = 6 ft
depth of water, d = 3 ft
discharge,Q = 250 ft³/s
now, we have to calculate Froude number
[tex]F_r = \dfrac{Q}{A\sqrt{gD}}[/tex]
Where D is the hydraulic radius
[tex]D = \dfrac{A}{P}[/tex]
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
P is the width of the channel
P = b + 2 z d
P = 6 + 2 x 1 x 3
P = 13 ft
A = d(b + z d) = 3 (6 + 3) = 27 ft²
g = 32.2 ft/s²
now,
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
[tex]F_r = \dfrac{250}{27\sqrt{32.2\times \dfrac{27}{13}}}[/tex]
F_r = 1.087
F_r > 1
Froude number is greater than 1 so, the flow is Super critical flow.
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch ay-t, vy-t, and y-t graphs for the motion.
a) Time at which velocity is +20.0 m/s: 2.04 s
b) Time at which velocity is -20.0 m/s: 6.12 s
c) Time at which the displacement is zero: t = 0 and t = 8.16 s
d) Time at which the velocity is zero: t = 4.08 s
e) i) ii) iii) The acceleration of the boulder is always [tex]9.8 m/s^2[/tex] downward
f) See graphs in attachment
Explanation:
a)
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
[tex]a=g=-9.8 m/s^2[/tex]
downward (acceleration due to gravity). So, we can use the following suvat equation:
[tex]v=u+at[/tex]
where:
v is the velocity at time t
u = 40.0 m/s is the initial velocity
a=g=-9.8 m/s^2 is the acceleration
We want to find the time t at which the velocity is
v = 20.0 m/s
Therefore,
[tex]t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s[/tex]
b)
In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when
v = -20.0 m/s
(the negative sign means downward)
We use again the suvat equation
[tex]v=u+at[/tex]
And substituting
u = +40.0 m/s
a=g=-9.8 m/s^2
We find the corresponding time t:
[tex]t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s[/tex]
c)
To solve this part, we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the displacement
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration
t is the time
We want to find the time t at which the displacement is zero, so when
s = 0
SUbstituting into the equation and solving for t,
[tex]0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0[/tex]
which gives two solutions:
t = 0 (initial instant)
[tex]u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s[/tex]
which is the instant at which the boulder passes again through the initial position, but moving downward.
d)
To solve this part, we can use again the suvat equation
[tex]v=u+at[/tex]
where
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration
We want to find the time t at which the velocity is zero, so when
v = 0
Substituting and solving for t, we find:
[tex]t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s[/tex]
e)
In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.
If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude
[tex]F=mg[/tex]
where m is the mass of the boulder and [tex]g[/tex] the acceleration of gravity.
According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
Combining the two equations, we get
[tex]ma=mg\\a=g[/tex]
So, the acceleration of the boulder is [tex]g=9.8 m/s^2[/tex] downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).
f)
Find the three graphs in attachment:
- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s
- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.
- Acceleration-time graph: the acceleration is constant and it is [tex]-9.8 m/s^2[/tex], so this graph is a straight horizontal line.
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Two light bulbs, A and B, are connected to a 120-V outlet (a constant voltage source). Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Which light bulb has a greater filament resistance?
Answer:
Bulb A has a greater resistance.
Explanation:
Electric power (P) = V²/R
P = V²/R................ Equation 1
Where P = power, V = Voltage, R = Resistance.
Make R the subject of the equation
R = V²/P ................ Equation 2
For Bulb A,
Given: V = 120 V, P = 60 W.
Substitute into equation 2
R = 120²/60
R = 240 Ω
For bulb B
Given: V =120 V, P = 100 W.
Substitute into equation 2
R = 120²/100
R = 14400/100
R = 144 Ω
Hence Bulb A has a greater resistance.
bulb A has the greater filament resistance.
What is resistance?This can be defined as the ability of a conducting material to oppose the flow of electric current.
To know the bulb with the greatest filament resistance, we use the formula below.
Formula:
P = V²/R................. Equation 1making R the subject of the equation
R = V²/P............... Equation 2Where:
P = power of the bulbV = Outlet voltageR = Resistance of the filament.For Bulb A,
Given:
P = 60 WV = 120 VSubstitute these values into equation 1
R = (120²)/60R = 240 ohms.For Bulb B,
Given:
P = 100 W.V = 60 VSubstitute these values into equation 1
R = (120²)/100R = 144 ohms.Hence, From the above, it can be seen that bulb A has the greater filament resistance.
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Raindrops hitting the side windows of a car in motion often leave diagonal streaks even if there is no wind. Why? Is the explanation the same or different for diagonal streaks on the windshield?
Answer:
because of the raindrop velocity relative of the car has a vertical and horizontal component
Explanation:
The car moves in a horizontal direction relative to the ground. The raindrops fall in the vertical direction relative to the ground. Their velocity relative to the moving car has both vertical and horizontal components and this is the reason for the diagonal streaks on the side window. The diagonal streaks on the windshield arise from a different reason. The drops are pushed off to one side of the windshield because of air resistance.A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C directed perpendicularly to the plate.a. Find the charge density of each face of the plate.b. Find the total charge on each face.
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
A hockey pick sliding along a frictional surface strikes a box at rest, after the collision the two objects stick together and move at the same final speed. Which of the following describes the change in momentum and energy of the puck during the collision?
a. puck loses some but not all of its original momentum.
b. one cannot determine
c. puck conserves original momentum, but loses all mechanical energy
d. puck loses some momentum but conserves mechanical energy
e. puck loses conserves all momentum and mechanical energy
f. conserves momentum but loses some mechanical energy
Answer:
Explanation:
Option a is correct
If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .
Option e is correct
If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.
Friction will cause the energy to dissipate in heat.
You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.2.2 m/s25.5 m/s23.7 m/s24.3 m/s2
Answer:
3.72 m/s²
Explanation:
Horizontal component of the force applied = Fcosθ = 86 cos 30 = 74.478 N
Force = mass × acceleration = 74.478 N
ma = 74.478 N
a = 74.478 / 20 since friction can be neglected = 3.72 m/s²
Answer:
[tex]3.7m/s^{2}[/tex]
Explanation:
The forces acting on the cart is displayed in the attached file below.
Since the motion is only along the horizontal, we can conclude that the force bringing about that motion is the horizontal force.And from second newton law of motion, we deduce that
F=ma,
F=force, m=mass and a=acceleration
from the given question,
F=86N, mass,m=20kg,
Hence we can write the horizontal component of the force as
[tex]F_{X}=Fcos\alpha \\F_{x}=86cos30\\F_{x}=74.48\\Hence \\a=\frac{F_{x}}{m} \\a=\frac{74.48}{20}\\ a=3.7m/s^{2}[/tex]
An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (a) Calculate the object’s position and acceleration as functions of time. (b) What is the object’s maximum positive displacement from the origin?
Answer:
Explanation:
Given
[tex]v_x(t)=\alpha -\beta t^2[/tex]
[tex]\alpha =4\ m/s[/tex]
[tex]\beta =2\ m/s^3[/tex]
[tex]v_x(t)=4-2t^2[/tex]
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\int dx=\int \left ( 4-2t^2\right )dt[/tex]
[tex]x=4t-\frac{2}{3}t^3[/tex]
acceleration of object
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]a=-4t[/tex]
(b)For maximum positive displacement velocity must be zero at that instant
i.e.[tex]v=0[/tex]
[tex]4-2t^2=0[/tex]
[tex]t=\pm \sqrt{2}[/tex]
substitute the value of t
[tex]x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}[/tex]
[tex]x=3.77\ m[/tex]
The definitions of acceleration and velocity allow to find the results for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
Given parameters
The velocity of the body v = α-β t² with α = 4 m/s and β = 2 m/s²To find
a) position and acceleration as a function of time,
b) maximum displacement,
The acceleration of defined as the change in velocity with time.
a = [tex]\frac{dv}{dt}[/tex]
Let's calculate.
a = - 2βt
a = - 2 2 t
a = -4 t
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
We integrate.
∫ dx = ∫ v dt
x - x₀ = ∫ (α - β t²)
x-x₀ = αt - βt³/ 3
we substitute.
x = 4 t - ⅔ t³
B) To find the maximum displacement we use the first derivative to be zero.
[tex]\frac{dx}{dt}[/tex] = 0
4 - 2t² = 0
t² = 2
t = √2 = 1.414 s
Let's find the position for this time.
x = 4 √2 - ⅔ (√2)³
x = 3.77 m
In conclusion using the definitions of acceleration and velocity we can find the result for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
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A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, find the initial velocity of the projectile. Neglect air resistance.
Answer:
66.02m/s
Explanation:
the equation describing the distance covered in the horizontal direction is
[tex]x=ucos\alpha t-(1/2)gt^{2}[/tex] but the acceleration in the horizontal path is zero, hence we have
[tex]x=ucos\alpha t[/tex]
Since the horizontal distance covered is 155m at 7.6secs, we have [tex]ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1[/tex]
Also from the vertical path, the distance covered is expressed as
[tex]y=usin\alpha t-(1/2)gt^{2}[/tex]
since the horizontal distance covered in 7.6secs is 195m, then we have
[tex]y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2[/tex]
Hence if we divide both equation 1 and 2 we arrive at
[tex]\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\[/tex]
Hence if we substitute the angle into the equation 1 we have
[tex]ucos72.02=20.38\\u=66.02m/s[/tex]
Hence the initial velocity is 66.02m/s
In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))
Answer:
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
Explanation:
Given that
g₁ = 9.96 m/s²
g₂ = 9.72 m/s²
The actual value of g = 9.8 m/s²
a)
The difference Δ g = 9.96 -9.72 =0.24 m/s²
[tex]The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\[/tex]
b)
For first one :
[tex]Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage[/tex]
For second :
[tex]Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage[/tex]
c)
The mean g(mean )
[tex]g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2[/tex]
[tex]The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage[/tex]
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.
Explanation:In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:
|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%
The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:
For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%
For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%
The percent error of the mean involves doing the above but using the mean of the experimental measurements:
|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%
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What is the final volume of a balloon that was initially 500.0 mL at 25°C and was then heated to 50°C?
Answer:
V₂ =541.94 m L.
Explanation:
Given that
V₁ = 500 mL
T₁ = 25°C = 273 + 25 = 298 K
T₂ = 5°C = 273+50 =323 K
The final volume = V₂
We know that ,the ideal gas equation
If the pressure of the gas is constant ,then we can say that
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]
The final volume of the balloon will be 541.94 m L.
V₂ =541.94 m L.
The final volume of the balloon will be "541.94 mL".
Given:
Volume,
[tex]V_1 = 500 \ mL[/tex]Temperature,
[tex]T_1 = 25^{\circ} C[/tex][tex]= 273+25[/tex]
[tex]= 298 \ K[/tex]
[tex]T_2 = 5^{\circ} C[/tex][tex]=273+50[/tex]
[tex]=323 \ K[/tex]
By using the Ideal gas equation, we get
→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]
or,
→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]
[tex]= \frac{500\times 323}{298}[/tex]
[tex]= 541.94 \ mL[/tex]
Thus the above approach is correct.
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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr = 8 gm, and cr = 3.7 J/gm ∘K-1. If the water is initially at room temperature, how long will it take for the water to heat up 5∘K? (Hint: dT/dt is approximately equal to Δ T / Δt .)
Answer:
[tex]t=5057.9167\ s[/tex]
Explanation:
Given:
Voltage supply to the resistor, [tex]V=24\ V[/tex]current supply to the resistor, [tex]I=0.1\ A[/tex]mass of water, [tex]m_w= 51\ g[/tex]specific heat of water, [tex]c_w=4180\ J.kg^{-1}.K^{-1}[/tex]specific heat of resistor, [tex]c_r=3700\ J.kg^{-1}.K^{-1}[/tex]mass of resistor, [tex]m_r=0.008\ kg[/tex]change in temperature, [tex]\Delta T=50\ K[/tex]Now the amount of heat required to heat the water by 50 K:
[tex]Q_w=m_w.c_w.\Delta T[/tex]
[tex]Q_w=0.051\times 4180\times 50[/tex]
[tex]Q_w=10659\ J[/tex]
Now the amount of heat required to heat the resistor by 50 K:
[tex]Q_r=m_r.c_r.\Delta T[/tex]
[tex]Q_r=0.008\times 3700\times 50[/tex]
[tex]Q_r=1480\ J[/tex]
Now the total heat to converted from the electrical energy:
[tex]Q=Q_w+Q_r[/tex]
[tex]Q=12139\ J[/tex]
Now Using Joule's law of heating:
[tex]Q=V.I.t[/tex]
[tex]12139=24\times 0.1\times t[/tex]
[tex]t=5057.9167\ s[/tex]
Final answer:
To determine the time needed to heat water by 5°K, calculate the total energy needed using the specific heat capacity of water, then use the power equation P = V × I to find the time. It will take approximately 444.125 seconds for the water to reach the desired temperature.
Explanation:
To calculate the time it will take for the water to heat up by 5°K using the provided values, first determine the total energy required to raise the temperature of the water by using the equation E = mw × cw × ΔT. Here, E is the energy in joules (J), mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given figures:
E = 51 g × 4.18 J/g°K × 5°K = 1065.9 J
Since power P is the rate at which energy is used and P = V × I where V is the voltage and I is the current, we can rearrange the equation to find time: t = E / P.
Substitute the power using the given voltage and current:
P = 24 V × 0.1 A = 2.4 W
The time in seconds to heat the water 5°K is thus:
t = 1065.9 J / 2.4 W = 444.125 seconds
Therefore, it will take approximately 444.125 seconds to heat up the water by 5°K.
Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced from 3.25 lb of azurite?
Answer:
1402.73 m
Explanation:
Mass of Azurite=3.25 lb
Percent of copper in AZurite mineral=55.1%
Diameter of copper wire,d=0.0113 in
Radius of copper wire=[tex]r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in[/tex]
[tex]\frac{1}{1000}=10^{-3}[/tex]
Density of copper=[tex]\rho=8.96g/cm^3[/tex]
1 lb=454 g
3.25 lb=[tex]3.25\times 454=1475.5 g[/tex]
Mass of Azurite=[tex]1475.5 g[/tex]
Mass of copper=[tex]\frac{55.1}{100}\times 1475.5=813 g[/tex]
Density=[tex]\frac{Mass}{volume}[/tex]
Using the formula
[tex]8.96=\frac{813}{volume\;of\;copper}[/tex]
Volume of copper wire=[tex]\frac{813}{8.96}=90.7cm^3[/tex]
Radius of copper wire=[tex]5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm[/tex]
1 in=2.54 cm
Volume of copper wire=[tex]\pi r^2 h[/tex]
[tex]\pi=3.14[/tex]
Using the formula
[tex]90.7=3.14\times (14.35\times 10^{-3})^2\times h[/tex]
[tex]h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}[/tex]
[tex]h=140273 cm[/tex]
1 m=100 cm
[tex]h=\frac{140273}{100}=1402.73 m[/tex]
Hence, the length of copper wire required=1402.73 m