Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s (88 mi/h). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.
Answer:
44.1613858478 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 99.4
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s[/tex]
If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s
When the skydiver descends to a certain height above the ground, she deploys her parachute to ensure a safe landing. Usually the parachute is deployed when the skydiver reaches an altitude of about 900 m (3000 ft). Immediately after deploying the parachute, does the skydiver have a nonzero acceleration?
Final answer:
After deploying the parachute, a skydiver does experience nonzero acceleration as the increased drag causes rapid deceleration until reaching a new lower terminal velocity.
Explanation:
Immediately after deploying her parachute, a skydiver does indeed have a nonzero acceleration. When a parachute opens, it rapidly increases the area exposed to air resistance, causing a significant increase in drag. As a result, the force of air resistance acting on the skydiver greatly exceeds the force of gravity, which causes a rapid deceleration of the skydiver. The magnitude of this deceleration is determined by the net force acting on the skydiver, which now includes the strong opposing force from the parachute's drag. This deceleration continues until the skydiver reaches a new, much lower terminal velocity with the parachute open.
When you try to stretch a bungee cord a distance x, it resists with an opposing force of the form b x2 , where b is a constant. If b is measured to be 6 N/m2 , how much work does it take to stretch the bungee cord a distance of 1 meterA.2JB.3JC.0.5JD. 1J
When a bungee cord is stretched a distance x, it exerts an opposing force given by the formula F = bx2. Integrated over the distance of stretch, this results in a work done of 2 Joules for a stretch distance of 1 meter and force constant b = 6 N/m^2.
Explanation:The work done in stretching an elastic object like a bungee cord is given by the integral of the force over the distance of stretch. Given the force exerted by the bungee cord is F = bx^2, where b = 6 N/m^2 and the work W is defined as the integral from 0 to x of F dx, the formula to calculate work done becomes W = ∫from 0 to 1 (bx^2) dx. Solving this integral gives W = (b/3)x^3 evaluated from 0 to 1, which simplifies to W = (6/3) (1) = 2 Joules. So, the correct answer is 2 Joules.
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Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface and push them together until they touch. Each ball has a mass of 4 kg. Think carefully about what to use for the distance between the two balls. And to keep things simpler, let's assume they slide without rolling.Now consider the case where both balls are free to move once you release them. Neglecting friction, calculate the final speed of one of the balls.?
Explanation:
First, we will calculate the electric potential energy of two charges at a distance R as follows.
R = 2r
= [tex]2 \times 0.1 m[/tex]
= 0.2 m
where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.
U = [tex]\frac{k \times Q \times Q}{R}[/tex]
= [tex]\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}[/tex]
= 0.081 J
As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.
Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, [tex]\frac{U}{2}[/tex].
Hence, K.E = [tex]\frac{U}{2}[/tex]
= [tex]\frac{0.081}{2}[/tex]
= 0.0405 J
Now, we will calculate the speed of balls as follows.
V = [tex]\sqrt{\sqrt{\frac{2 \times K.E}{m}}[/tex]
= [tex]\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}[/tex]
= 0.142 m/s
Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.
A boy throws a rock with an initial velocity of at 30.0° above the horizontal. If air resistance is negligible, how long does it take for the rock to reach the maximum height of its trajectory?
Answer: 0.5m/s
This value may vary depending on the initial velocity value.
Explanation:
The question is incomplete due to absence of the initial velocity (U) of the rock. Taking the initial velocity as any value say 10m/s at 30° above the horizontal.
Time it will take to reach maximum height (Tmax) will be Usin(theta)/g
Where U is the initial velocity = 10m/s theta = 30° g is the acceleration due to gravity = 10m/s²
Substituting the values in the formula we have;
Tmax = 10sin30°/10
Tmax = sin 30°
Tmax = 0.5second
Therefore, it will take for the rock 0.5s to reach the maximum height of its trajectory if the velocity is 10m/s.
The time varies though depending on the value of the initial velocity.
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far along the length of the ladder can a 69.1-kg painter climb before the ladder begins to slip?
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=[tex]\frac{N2}{mg}*l[/tex] * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= [tex]\frac{241.75}{69.1*9.8} *l[/tex] * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
The weight of the ladder is given as well as the length ; l = 11.9 m long and weighing W = 46.1 N rests against a smooth vertical wall.
how far along the length of the ladder = x = 4.14m
Explanation:
The detailed step and calculation is as shown in the attached file.
The acceleration of a particle varies with time according to the equation a(t) = pt^2 - qt^3. Initially, the velocity and position are zero, (a) What is the velocity as a function of time? (b) What is the position as a function of time?
Answer:
Explanation:
Given
acceleration of a particle is given by [tex]a(t)=pt^2-qt^3[/tex]
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=pt^2-qt^3[/tex]
[tex]dv=(pt^2-qt^3) dt[/tex]
Integrating we get
[tex]\int dv=\int \left ( pt^2-qt^3\right )dt[/tex]
[tex]v=\frac{pt^3}{3}-\frac{qt^4}{4}+c_1[/tex]
at [tex]t=0\ v=0[/tex]
therefore [tex]c_1=0[/tex]
We know velocity is given by
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]vdt=dx[/tex]
integrating
[tex]\int dx=\int \left ( \frac{pt^3}{3}-\frac{qt^4}{4}+\right )dt[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}+c_2[/tex]
using conditions
at [tex]t=0\ x=0[/tex]
[tex]c_2=0[/tex]
[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}[/tex]
(a) The velocity of the particle as function of time is pt³/3 - qt⁴/4 + C.
(b) The position of the particle as function of time is pt⁴/12 - qt⁵/20 + Ct + C.
Velocity of the particleThe velocity of the particle is the integral of the acceleration of the particle.
The velocity of the particle is calculated as follows;
v = ∫a(t)
v = ∫(pt² - qt³)dt
v = pt³/3 - qt⁴/4 + C
Position of the particleThe position of the particle as function of time is the integral of velocity of the particle.
x = ∫v
x = ∫(pt³/3 - qt⁴/4 + C)dt
x = pt⁴/12 - qt⁵/20 + Ct + C
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Colossal Cave was formed years ago by underground running water. Today, it is the home to many animals, like bats. These interactions are an example of _____. a. biosphere, atmosphere, and lithosphere b. interaction hydrosphere, lithosphere, and biosphere c. interaction hydrosphere, lithosphere, and atmosphere d. interaction atmosphere and biosphere interaction
Answer:
d. interaction atmosphere and biosphere interaction
Explanation:
hydrosphere, lithosphere, and biosphere interaction.
Apply Newton's first law to each of the following situations. In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? A book slides across the table and comes to a stop.
Answer:
kinetic frictional force opposes the relative motion between the surfaces in contact of the book and the table.
Explanation:
When a book slides on across the table and comes to stop then there must be force acting on it which hinder its state of uniform motion.According to the Newton's first law of motion every body continues to be in the state of rest or in uniform motion until acted upon by any external force.Here while the book slides on the table there acts a force of friction between the table surface and the surface of the book which is in contact to the top of the table while the relative motion between the surfaces there acts a kinetic frictional force which opposes the relative motion between the two.2 Grade 11 Physics Questions. Will Mark BRAINLIEST!!!
Which machine would be an alternative to lifting a crate with a pulley?
Ramp
Screw
Doorstop
Ax
Which of the following is an example of a simple machine?
Bicycle
Car
Steering wheel system of a boat
Ramp
Answer:
For the first one it would be screw and for the second one it is a ramp
Explanation:
Hope this helps
Frequency division multiplexing: A. operates by statistically time slicing the signal B. operates by dividing the signal into different frequencies C. uses a codec that divides signals into different channels D. operates by time slicing the signal operates by light dividing the signal
Frequency division multiplexing operates by dividing the signal into different frequencies
Explanation:
The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.
Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.
Frequency Division Multiplexing operates by dividing a signal into different frequencies to enable multiple transmissions at the same time. This method is used in FM radio and television broadcasts, as well as cell phone conversations and computer data transmissions.
Explanation:Frequency Division Multiplexing (FDM) operates by dividing the signal into different frequencies. This can be seen in how FM (Frequency Modulation) radio signals are used. In FM radio transmission, the information is carried by varying the frequency of the carrier wave and keeping its amplitude constant. This forms different channels on which multiple signals can be transmitted simultaneously without interfering with each other.
Another example is how television broadcasts. Since a vast amount of visual and audio information needs to be carried, each channel requires a larger range of frequencies, these fall under VHF and UHF (high to ultra-high frequencies).
Cell phone conversations and computer data are also transmitted using a similar approach by converting the signal into a sequence of binary ones and zeros. This binary sequence can be transmitted via different frequencies, allowing multiple data transmissions to take place at the same time.
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Two parallel conducting plates that are in deep space are brought to a potential difference of 3000 V, and a small pellet of mass 2.60 mg carrying a charge of 6.00 × 10-7 C accelerates from rest from the positive plate. With what speed will it reach the other plate?
Answer:
qU = Ek = ½mv²
6.00 * 10^-7 * 3000 = ½ * 0.0026 * v²
v² = 0.7222167
v = 0.8498 m/s
A park merry-go-round consists of a 255 kg circular wooden platform 4.10 m in diameter.?
Four children running alongside push tangentially along the platform's circumference until, starting from rest, the merry-go-round reaches a steady speed of one complete revolution every 2.8 s.
(a) If each child exerts a force of 26 N, how far does each child run?
(b) What is the angular acceleration of the merry-go-round?
(c) How much work does each child do?
(d) What is the kinetic energy of the merry-go-round?
Answer: a) 12.9 m b) 0.40 rad/s² c) 335.4 J d) 1349.06 J
Explanation:
a)
Distance traveled= R∅
where ∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α
To find α we have
Sum of torque= Iα
F x r x number of children = m*r²/2 * α
26 x [tex]\frac{4.1}{2}[/tex] * 4 = (255) * (4.10/2)² /2 *α
Solving we get,
α= [tex]\frac{213.2}{535.82}[/tex]
α= 0.40
So,
∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α
Becomes
∅= 6.29
Now,
Distance traveled= 4.1/2 * 6.29
Distance traveled= 12.9 m
b)
Angular acceleration= 0.40
c)
Work done= F x distance traveled
Work done= 26 x 12.9
Work done= 335.4 J
d)
Kinetic energy= [tex]\frac{1}{2}* I * (\frac{2\pi }{2.8})^2[/tex]
Putting values we get,
Kinetic energy= 267.91 * 5.03
Kinetic energy= 1349.06 J
The merry-go-round problem involves calculating the distance each child runs, the angular acceleration, the work done by each child, and the kinetic energy of the system, using principles of rotational dynamics and energy.
Explanation:We are given a merry-go-round scenario with children exerting force to set it in motion. Let's break down the problem step by step.
(a) Distance run by each child:
Determine the circumference of the merry-go-round: Circumference = π × diameter = 3.14159 × 4.10 m.Find the number of revolutions: One revolution takes 2.8 seconds, so the number of revolutions is the time spent pushing divided by 2.8 s.Distance run by each child is the product of the number of revolutions and the circumference.(b) Angular acceleration:
Calculate angular velocity (omega), given by omega = 2π / T, where T is the period (2.8 s).Use the kinematic equation for rotational motion: omega = omega_initial + alpha × t, where omega_initial = 0 (starting from rest) and t is the time. Solve for alpha (angular acceleration).(c) Work done by each child:
Work done equals force times distance: Work = force × distance run by the child.(d) Kinetic energy of the merry-go-round:
Find moment of inertia (I) for the merry-go-round: I = (1/2) × mass × radius^2.Calculate kinetic energy: KE = (1/2) × I × omega^2.
A laser dazzles the audience in a rock concert by emitting green light with a wavelength of 515 nm . Calculate the frequency of the light. Express your answer with the appropriate units.
Answer:
Explanation:
This question is quite tricky. Not all parameters were provided.
λ = 515nm(nm = namo meters), 515 nm becomes 515 x 10^-9 which in turn becomes 5.15 x 10^-7.
The velocity for the green light wasn't provided but the velocity of greenlight is a constant; velocity of green light(v) = 3.00 × 108 m/s.
frequency of the light =[tex]\frac{velocity of green light}{wavelength of light}\\[/tex]
f =[tex]\frac{3.00 X 108 m/s}{5.15 X 10^{-7} }[/tex]
f=20.971 Hertz
The frequency of light will be "20.971 Hz".
According to the question,
Wavelength,
[tex]\lambda = 515 \ nm[/tex]or,
[tex]= 5.15\times 10^{-7} \ m[/tex]
Velocity of green,
[tex]v = 3.00\times 108 \ m/s[/tex]The frequency of light (f) will be:
= [tex]\frac{Velocity}{Wavelength}[/tex]
= [tex]\frac{3.00\times 108}{5.15\times 10^{-7}}[/tex]
= [tex]20.971 \ Hz[/tex]
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An 80-kg skater is coasting at a velocity of 6 m/s. She sees a small child in her way and picks him up as she skates by. Her velocity decreases to 1.12 m/s. What is the mass of the child she picked up?
Answer:
348.57 kg
Explanation:
From the law of conservation of momentum,
Total momentum before the small child was picked = Total momentum after the small child was picked
mu + m'u' = V(m+m')................... Equation 1
Where m = mass of the skater, u = initial velocity of the skater, m' = mass of the small child, u' = initial velocity of the small child, V = common velocity after the child was picked.
Note: Assuming the small child was stationary before he was picked, the u' = 0 m/s.
making m' the subject of the equation
m' = (mu-mV)/V........................... Equation 2
Given: m = 80 kg, u = 6 m/s, V = 1.12 m/s.
Substitute into equation 2
m' = [(80×6)-(80×1.12)]/1.12
m' = (480-89.6)/1.12
m' = 390.4/1.12
m' = 348.57 kg.
Thus the mass of the small child = 348.57 kg
a 8.0 * 104 kg spaceship is at rest in deep space. it's thrusters provide a force of 1200 kn. the spaceship fires its thrusters for 20 s, then coasts for 12 km. how long does it take the spaceship to coast this distance?
Answer:
40 seconds
Explanation:
Thrusting speed = V
Where V=at
a= F/m
Therefore V = Ft/m
F = 1200kN = 1200000N
t = 20s
m = 8.0 * 104 kg
V = 1200000 * 20 / 80000
V = 300m/s
Time for the spaceship to coast a distance of 12km
Distance = Vt
t = 12000/300 = 40seconds
The time taken for taking the spaceship to coast this distance is 40 seconds
The calculation is as follows:
Thrusting speed = V
Where
V=at
And,
[tex]a= F\div m[/tex]
Therefore [tex]V = Ft\div m[/tex]
Now
F = 1200kN
= 1200000N
t = 20s
[tex]m = 8.0 \times 104 kg[/tex]
Now
[tex]V = 1200000 \times 20 \div 80000[/tex]
V = 300m/s
Time for the spaceship to coast a distance of 12km
Distance = Vt
[tex]t = 12000\div 300[/tex]
= 40seconds
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If A, B, and C are three activities connected with SS and FF (combination) relationships with 2-day lag on SS relationship only. Their durations are 5, 10, and 3 respectively. What is the duration of the project?a. 5 days. b. 10 days. c. 3 days. d 18 days.
Answer:
B. 10 days
Explanation:
Which of the following must ALWAYS be equal to the buoyant force on an object?
A.the force of gravity
B.the weight of the object
C.the weight of the liquid displaced by the objec
D.the force of friction as the object moves in the water
Answer:
D.
Explanation:
It just makes sense. It is a simple answer
Two small, identical particles have charges Q1 = +5.5 μC and Q2 = -17μC. The particles are conducting and are brought together so that they touch. Charge then moves between the two particles so as to make the excess charge on the two particles equal. If the particles are then separated by a distance of 60 mm, what is the magnitude of the electric force between them?
Answer:
Explanation:
Given
Charge [tex]Q_1=+5.5\mu C[/tex]
[tex]Q_2=-17\ mu C[/tex]
When the two charges are touch then the net charge on the two sphere will be same after some time
[tex]Q=\frac{Q_1+Q_2}{2}[/tex]
[tex]Q=\frac{5.5-17}{2}=-5.75\ \mu C[/tex]
Now the force between this tow charges are when they are separated with [tex]d=60\ mm[/tex]
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]F=\frac{9\times 10^9\times (-5.75)^2}{(60\times 10^{-3})^2}[/tex]
[tex]F=82.65\ N[/tex]
A 0.10-kilogram ball dropped vertically from a height of 1.0 meter above the floor bounces back to a height of 0.80 meter. The mechanical energy lost by the ball as it bounces is approximately
A) 0.20 J
B) 0.080 J
C) 0.78 J
D) 0.30 J
Answer:
A) 0.20Joules
Explanation:
The type mechanical energy acting on the body is potential energy since the body is covering a particular height.
Potential Energy = mass×acceleration due to gravity × height
Mass of the body = 0.1kg
acceleration due to gravity = 10m/s²
h is the height of the object when dropping = 1.0meters
Substituting this values in the formula to get the energy of the body on dropping, we have;
PE = 0.1×10×1.0
PE = 1.0Joules
On bouncing back to height of 0.8m, the potential energy becomes
PE = 0.1×10×0.8
PE = 0.8Joules
The mechanical energy lost by the ball as it bounces will approximately be the difference in its potential energy when dropping and when it bounces back i.e 1.0Joules - 0.8Joules = 0.20Joules
0.2 J
mgh (before) = (0.1)(9.8)(1) = 0.98
mgh (after) = (0.1)(9.8)(0.8) = 0.784
0.98-0.784=0.196
0.196 is approximately 0.2.
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b. The positive charge per unit length on the inner cylinder is λ, and there is an equal negative charge per unit length on the outer cylinder /a b, (iii) r > b Calculate the potential V(r) for (i) r ca,俪) a (Hint: The net potential is the sum of the potentials due to the individual conductors.) Take V 0 at r-b. a. 2charge per unit length lm上 b. Show that the potential of the inner cylinder with respect to the outer is -I c. Show that the electric field at any point between the cylinders has magnitude
a)
i) Potential for r < a: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
ii) Potential for a < r < b: [tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
iii) Potential for r > b: [tex]V(r)=0[/tex]
b) Potential difference between the two cylinders: [tex]V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c) Electric field between the two cylinders: [tex]E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
Explanation:
a)
Here we want to calculate the potential for r < a.
Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is
[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]
where
[tex]\lambda[/tex] is the linear charge density
r is the distance from the wire/surface of the cylinder
By integration, we find an expression for the electric potential at a distance of r:
[tex]V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)[/tex]
Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:
[tex]E=0[/tex]
So the potential where the electric field is zero is constant:
[tex]V=const.[/tex]
iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density [tex]+\lambda[/tex] and an equal negative charge density [tex]-\lambda[/tex]. Therefore, the net charge is zero, so the electric field is zero.
This means that the electric potential is constant, so we can write:
[tex]\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)[/tex]
However, we know that the potential at b is zero, so
[tex]V(r)=V(b)=0[/tex]
ii) The electric field in the region a < r < b instead it is given only by the positive charge [tex]+\lambda[/tex] distributed over the surface of the inner cylinder of radius a, therefore it is
[tex]E=\frac{\lambda}{2\pi r \epsilon_0}[/tex]
And so the potential in this region is given by:
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex] (1)
i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):
E = 0
This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
And so, for r<a,
[tex]V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
b)
Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.
We have:
- Potential at the surface of the inner cylinder:
[tex]V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
- Potential at the surface of the outer cylinder:
[tex]V(b)=0[/tex]
Therefore, the potential difference is simply equal to
[tex]V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})[/tex]
c)
Here we want to find the magnitude of the electric field between the two cylinders.
The expression for the electric potential between the cylinders is
[tex]V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}[/tex]
The electric field is just the derivative of the electric potential:
[tex]E=-\frac{dV}{dr}[/tex]
so we can find it by integrating the expression for the electric potential. We find:
[tex]E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}[/tex]
So, this is the expression of the electric field between the two cylinders.
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Which seismic wave is characterized by alternating compression-expansion parallel to the direction of wave movement?
Answer: Acoustic or sound wave
Explanation:
Acoustic wave is a type of wave energy that travels through a medium by adiabatic compression and decompression, they have acoustic velocity which is determined by the type and nature of medium they travel through. They are mechanical and longitudinal waves with characteristic features such as amplitude, period, frequency and wavelength.
A car traveling at 60 mph has how much more energy than a car going at 15 mph?
KE increases by a factor of____.
1. 1/16
2. 4
3. 1/4
4. 16
4. KE increases by a factor of 16 is the answer
Explanation:
Kinetic energy = (1/2)mv² = 0.5 mv²
where
m = mass, and v = velocity.
So at 15 mph,
K E = 0.5 m (15) ² = 112.5 m
And at 60 mph,
K E = 0.5 m (60)² = 1800 m
m is the mass, and not meters.
So, 1800 m/112. 5 m = 16
16 times the Kinetic Energy.
The drawings show three situations in which a positively charged particle is moving through a uniform magnetic field B with a velocity v. For each situation, what is the direction of the magnetic force F exerted on the particle?
Answer:
Case A: - x axis
Case B: - x axis
Case C: no force as the direction of v and B are parallel.
Explanation:
Solution:
- The direction of Force exerted by the magnetic field B with a moving charge q with a velocity of v is given by an expression:
F = q*(v x B)
- Where F: force vector exerted by the field
v: velocity vector of charged particle q.
B: magnetic field direction
- The cross product of v with B:
We will use right hand rule in which our fingers curl from v to B, the direction of the thumb denotes the direction of applied force. Hence,
Case A: - x axis
Case B: - x axis
Case C: no force as the direction of v and B are parallel.
Final answer:
The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). The force is perpendicular to the plane formed by the particle's velocity and the magnetic field. The direction of the force depends on the charge of the particle and is opposite for positive and negative charges.
Explanation:
The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). According to RHR-1, if you point your right thumb in the direction of the particle's velocity (v) and your right fingers in the direction of the magnetic field (B), then your right palm will point in the direction of the magnetic force (F) acting on the particle.
For example, if the particle is moving perpendicular to the magnetic field (B), as shown in the figure, and the particle has a positive charge, then the force (F) will be directed outward from the plane formed by v and B.
It's important to note that the direction of the magnetic force on a negatively charged particle will be in the opposite direction to that on a positively charged particle.
Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newton Nicolaus Copernicus Galileo Galilei
Answer:
Option (3)
Explanation:
Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.
He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.
This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.
Thus, the correct answer is option (3).
Answer:
I think it is option 'c' in the quiz.
Explanation:
I took the test and got a 100.
The vibrations along a transverse wave move in a direction _________.
Answer: perpendicular to it oscillations.
Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.
By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.
Examples of transverse waves includes wave in a string, water wave and light.
Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.
If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.
Since the oscillations is perpendicular to the direction of wave, it is a transverse wave
In a transverse wave, the vibrations move in a direction perpendicular to the direction of wave travel. This distinguishes them from other types of waves. An example is the movement of water waves.
Explanation:In a transverse wave, the vibrations or disturbances move in a direction perpendicular (at right angles) to the direction of the wave's travel. This is the key characteristic that separates transverse waves from other types of waves. A common example of transverse waves would be waves in a body of water where the water particles move up and down causing the wave to move forward.
Learn more about transverse wave here:https://brainly.com/question/33916729
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Which is true about rule utilitarianism and Kantianism (check all that apply) a. Both theories evaluate the moral rules in the same way b. According to both theories, right actions are those that are in line with universal moral rules c. Both are consequentialist theories d. Both are relativistic theories
Answer:
A
Explanation:
Why is that they truly evaluate both of the theories in the same way.
An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and its volume is 1.08×1012km3. What is the earth’s average density?
Final answer:
To find the Earth's average density, we divide its mass (5.94×10²⁴ kg) by its volume converted to cubic meters (1.08×10²⁴ m³), resulting in a density of approximately 5.5 kg/m³.
Explanation:
An object's average density is defined as the ratio of its mass to its volume (rho = M/V). To calculate the Earth's average density, we use the given mass of the Earth (5.94×10²⁴ kg) and the given volume of the Earth (1.08×10¹² km³).
It's important to convert the volume from cubic kilometers to cubic meters to ensure consistency in units since density is typically expressed in kilograms per cubic meter (kg/m³). There are 1,000,000,000,000 (1×10¹²) cubic meters in one cubic kilometer, so the volume in cubic meters is 1.08×10¹² km³ × 1×1012 m³/km³ = 1.08×1024 m³.
Now, dividing the mass by the volume in consistent units gives:
Density = Mass / Volume = 5.94×10²⁴ kg / 1.08×10²⁴ m³ = 5.5 kg/m³
Therefore, The average density of Earth is roughly 5.5 kg/m³.
What is the velocity of an electron that has a de Broglie wavelength approximately the length of a chemical bond? Assume this length to be 1.2 * 10-10m.
Answer : The velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]
Explanation :
According to de-Broglie, the expression for wavelength is,
[tex]\lambda=\frac{h}{p}[/tex]
and,
[tex]p=mv[/tex]
where,
p = momentum, m = mass, v = velocity
So, the formula will be:
[tex]\lambda=\frac{h}{mv}[/tex] .............(1)
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\lambda[/tex] = wavelength = [tex]1.2\times 10^{-10}m[/tex]
m = mass of electron = [tex]9.11\times 10^{-31}kg[/tex]
v = velocity of electron = ?
Now put all the given values in formula 1, we get:
[tex]1.2\times 10^{-10}m=\frac{6.626\times 10^{-34}Js}{(9.11\times 10^{-31}kg)\times v}[/tex]
[tex]v=6.1\times 10^{6}m/s[/tex]
Thus, the velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]
Humans have existed for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as 1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existed?
Answer:
8.64 seconds
Explanation:
According to the question
[tex]10^{10}\ years=1\ day[/tex]
[tex]1\ year=\dfrac{1}{10^{10}}\ day[/tex]
Humans have lived for [tex]10^6[/tex] years
In one universe day
[tex]10^6\ years=10^6\times \dfrac{1}{10^{10}}=10^{-4}\ day[/tex]
In seconds
[tex]10^{-4}\times 24\times 60\times 60=8.64\ seconds[/tex]
In one universe day humans have lived for 8.64 seconds
Humans have been around for 8.64 'universe seconds'.
The question asks for a conversion of human existence into 'universe seconds' given that humans have been around for approximately 106 years and the universe is approximately 10^10 years old. To answer this, one must conceptualise a 'universe day' as representing the entire 10^10 years of the universe's existence and break it down into 'universe seconds' in the same way that a normal day is broken down into normal seconds.
A regular day has 86,400 seconds (24 hours × 60 minutes per hour × 60 seconds per minute). As humans have existed for 10^6 years in the context of the universe's 10^10 years, we can calculate the fraction of the universe's existence that humans have been present for:
Human existence fraction = (Human years) / (Universe years) = 10^6 / 10^10 = 1 / 10^4
Now, we multiply this fraction by the total number of seconds in a day to find the equivalent 'universe seconds' of human existence:
Universe seconds of human existence = 1 / 10^4 × 86,400
Universe seconds of human existence = 8.64
Therefore, humans have existed for 8.64 'universe seconds' when defining the age of the universe as 1 'universe day'.