Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above
the horizontal, Figure 1. A constant force of 18 N is applied on the box in a direction 10°
with the horizontal causing the box to accelerate up the incline.
The coefficient of
kinetic friction between the block and the plane is 0.25.
Show the free body diagrams
(a) Calculate the block's
acceleration as it moves up the incline. (6 marks)
(b) If the block slides down at a constant speed, find the value of force applied.
(4 marks)
Answer:
a) a = 17.1 m / s², b) F = 3.04 N
Explanation:
This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities
* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components
* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components
We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is
θ = 10 -30 = -20º
The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force
sin (-20) = F_{y} / F
cos (-20) = Fₓ / F
F_{y} = F sin (-20)
Fₓ = F cos (-20)
F_y = 18 sin (-20) = -6.16 N
Fₓ = 18 cos (-20) = 16.9 N
The decomposition of the weight is the customary
sin 30 = Wₓ / W
cos 30 = W_y / W
Wₓ = W sin 30 = mg sin 30
W_y = W cos 30 = m g cos 30
Wₓ = 0.8 9.8 sin 30 = 3.92 N
W_y = 0.8 9.8 cos 30 = 6.79 N
Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane
Now we can write Newton's second law for each axis
X axis
Fₓ - fr = m a
Y Axis
N - [tex]F_{y}[/tex] - Wy = 0
N =F_{y} + Wy
N = 6.16 + 6.79
They both go to the negative side of the axis and
N = 12.95 N
The friction force has the formula
fr = μ N
we substitute
Fₓ - μ N = m a
a = (Fₓ - μ N) / m
we calculate
a = (16.9 - 0.25 12.95) / 0.8
a = 17.1 m / s²
b) now the block slides down with constant speed, therefore the acceleration is zero
ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced
Newton's law for the x axis
Fₓ -fr = 0
Fₓ = fr
F cos 20 = μ N
F = μ N / cos 20
we calculate
F = 0.25 12.95 / cos 20
F = 3.04 N
this is the force applied at an angle of 10º to the horizontal
Light with a frequency of 5.70×10^14 Hz travels in a block of glass that has an index of refraction of 1.56. What is the wavelength of the light in the glass?
✔First you have to calculate the light's speed in the glass,
You know that in the air and in the void (where the refraction index n is zero) the light's speed C corresponds to 3,0 x 10^8 m/s
So We have :
V = C/n
V = 3,0 x 10^8/1,56 V ≈ 1,92 x 10^8 m/s✔ Now, you know the light's speed in glass, and you know that : the wavelength λ is the quotient of light's speed V on its frequency ν, so :
λ = V/ ν
λ = 1,82 x 10^8/5,70 x 10^14 λ ≈ 3.40 x 10^-7 m λ ≈ 340 nmwith a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.
Answer:
120000 kgxm/s
Explanation:
momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000 kgxm/s
which of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively charged objects attract negatively charged objects. Positively charged objects attract neutral objects. Negatively chargers objects attract neutral objects.
Answer:
negatively charged object attract other negatively objects
Explanation:
opposites attract
Answer:
negativelycharged objects attract other negatively charged objects
Explanation:
unlike charges attract like charges repel
The cost of buying shirts is partly constant and partly varies with the number of shirts bought. When the number of shirts is 5 the cost is #240, also, 10 shirts costs #400. find the cost when 300 shirts were bought
Answer:
The cost of the buying the shirts is #9680
Explanation:
let the cost of buying shirt = C
let the number of shirt bought = N
The following equation can be generated based on the statement above;
C = k + Nb
When the cost, C = #240, the number of shirt = 5
240 = k + 5b ------ equation (1)
where;
k and b are constants
When the cost, C = #400, the number of shirt = 10
400 = k + 10b ------ equation (2)
From equation (1), make k the subject of the formula;
k = 240 - 5b ---- equation (3)
Substitute in the value of k into equation (2)
400 = k + 10b
400 = (240 - 5b) + 10b
400 = 240 - 5b + 10b
400 - 240 = -5b + 10b
160 = 5b
b = 160 / 5
b = 32
From equation (3), calculate k
k = 240 - 5b
k = 240 -5(32)
k = 240 - 160
k = 80
When the number of shirts bought = 300, the cost of the buying the shirts =
C = k + Nb
C = 80 +32N
Where;
N is the number of shirts
C = 80 + 32(300)
C = 80 + 9600
C = #9680
Therefore, the cost of the buying the shirts is #9680
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
The resistance of the original circuit is [tex]32\,\,\Omega[/tex]
Explanation:
In the original circuit, we have an unknown resistor that we call R, an unknown power supply that we call V, and the current is 15 Amps. in the second circuit with an added 8 Ω resistor in series, which gives an equivalent resistance of R+8 Ω, using the same power supply V, the current is 12 Amps. SO, we can write a system of two equations with two unknowns as follows:
[tex]V=R\,(15)\\V=(R+8)\,(12)\\then\\15\,R=12\,R+92\\3\,R=96\\R=\frac{96}{3} \,\Omega\\R=32\,\,\Omega[/tex]
Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an incline, all with the same initial kinetic energy.
Which goes furthest the incline?
a. The hoop
b. The disk
c. The sphere
d. They all roll to the same height
Answer:
The sphere
Explanation:
Because it has a smaller inertia (I) value in the explanation in the attached file
Two capacitors, CA and CB, are such that CA > CB. These are connected with a battery in various ways: each individually, series, and parallel. Rank these four cases according to the total amount of charge, greatest first.
a. (CA) > (C) > (CA and CB in series) > (CA and Co in parallel)
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
c. (CA and CB in series) > (CA) > (CB) > (CA and CB in parallel)
d. (CA and Cg in parallel) > (CA) > (CB) > (CA and Cg in series)
Answer:
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
Explanation:
This is because capacitors in series is the sum of the reciprocal of the capacitances while that of parallel is the sum of the individual capacitances
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
Assume that a lightning bolt can be represented by a long straight line of current. If 15.0 C of charge passes by in a time of 1.5·10-3s, what is the magnitude of the magnetic field at a distance of 24.0 m from the bolt?
Answer:
The magnitude of the magnetic field is 8.333 x 10⁻⁷ T
Explanation:
Given;
charge on the lightening bolt, C = 15.0 C
time the charge passes by, t = 1.5 x 10⁻³ s
Current, I is calculated as;
I = q / t
I = 15 / 1.5 x 10⁻³
I = 10,000 A
Magnetic field at a distance from the bolt is calculated as;
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷
I is the current in the bolt
r is the distance of the magnetic field from the bolt
[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ????) of the system?
The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.
Given the following data:
Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]Work done = [tex]4.50 \times 10^2 \;kcal[/tex]Conversion:
1 kcal = 4.184 kJ
[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]
To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.
Mathematically, the first law of thermodynamics is given by the formula:
[tex]\Delta E = Q - W[/tex]
Where;
[tex]\Delta E[/tex] is the change in internal energy.Q is the quantity of heat released.W is the work done.Substituting the given parameters into the formula, we have;
[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]
Change in internal energy, E = -18823 Kilojoules
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Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you
Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE [tex]e^{-t/RC}[/tex]
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit
A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that the price of the car is $32,540. Why is the long shadow of scarcity visible at the car dealership? Check all that apply.
Answer:
he will see the sticker because its behind a window bruh and thats a big daddy stack of greens
Explanation:
A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 11.3 m/s after the collision. If they skid for 14.7 seconds before stopping, how far did they skid
Answer:
83.055 m
Explanation:
According to the given scenario, the calculation of skid distance is shown below:-
[tex]S = \frac{1}{2} \times (u + v) \times t[/tex]
Where
u = 11.3
v = 0
t = 14.7
Now placing these values to the above formula,
So,
[tex]S = \frac{1}{2} \times (11.3 + 0) \times 14.7[/tex]
= 83.055 m
Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answer:
Explanation:a 1
Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris wheel rotates once every 24 s.What is the apparent weight of a 40 kg passenger at the lowest point of the circle?What is the apparent weight of a 40 kg passenger at the highest point of the circle?
Answer:
a
[tex]F_A =425.42 \ N[/tex]
b
[tex]F_A_H = 358.58 \ N[/tex]
Explanation:
From the question we are told that
The diameter of the Ferris wheel is [tex]d = 80 \ ft = \frac{80}{3.281} = 24.383[/tex]
The period of the Ferris wheel is [tex]T = 24 \ s[/tex]
The mass of the passenger is [tex]m_g = 40 \ kg[/tex]
The apparent weight of the passenger at the lowest point is mathematically represented as
[tex]F_A_L = F_c + W[/tex]
Where [tex]F_c[/tex] is the centripetal force on the passenger, which is mathematically represented as
[tex]F_c =m * r * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2* \pi }{T}[/tex]
substituting values
[tex]w = \frac{2* 3.142 }{24}[/tex]
[tex]w = 0.2618 \ rad/s[/tex]
and r is the radius which is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{24.383}{2}[/tex]
[tex]r = 12.19 \ ft[/tex]
So
[tex]F_c = 40 * 12.19* (0.2618)^2[/tex]
[tex]F_c = 33.42 \ N[/tex]
W is the weight which is mathematically represented as
[tex]W = 40 * 9.8[/tex]
[tex]W = 392 \ N[/tex]
So
[tex]F_A = 33.42 + 392[/tex]
[tex]F_A =425.42 \ N[/tex]
The apparent weight of the passenger at the highest point is mathematically represented as
[tex]F_A_H = W- F_c[/tex]
substituting values
[tex]F_A_H = 392 - 33.42[/tex]
[tex]F_A_H = 358.58 \ N[/tex]
The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what direction must the lens be moved to change the focus of the camera to a person 4.0 m away
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.
Answer:
Zack should direct his throw outward and toward the back of the car.
Explanation:
As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.
The solution is throw 3.
I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Which statement best applies Newton’s laws of motion?The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.
When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.
The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.
Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
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Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determine:
(a) the magnitude and direction of the vector D=A+B+C and
(b) the magnitude and direction of E=-A - B + C.
Answer:
(a) [tex]\vec D = 2\,i - 2\,j[/tex], (b) [tex]\vec E = -6\,i + 12\,j[/tex]
Explanation:
Let be [tex]\vec A = 3\,i - 3\,j\,[m][/tex], [tex]\vec B = i - 4\,j\,[m][/tex] and [tex]\vec C = -2\,i + 5\,j \,[m][/tex], each resultant is found by using the component method:
(a) [tex]\vec D = \vec A + \vec B + \vec C[/tex]
[tex]\vec D = (3\,i - 3\,j) + (i-4\,j) + (-2\,i+5\,j)\,[m][/tex]
[tex]\vec D = (3\,i + i -2\,i)+(-3\,j-4\,j+5\,j)\,[m][/tex]
[tex]\vec D = (3 + 1 -2)\,i + (-3-4+5)\,j\,[m][/tex]
[tex]\vec D = 2\,i - 2\,j[/tex]
(b) [tex]\vec E = -\vec A - \vec B + \vec C[/tex]
[tex]\vec E = -(3\,i-3\,j)-(i - 4\,j)+(-2\,i+5\,j)[/tex]
[tex]\vec E = (-3\,i + 3\,j) +(-i+4\,j) + (-2\,i + 5\,j)[/tex]
[tex]\vec E = (-3\,i-i-2\,i) + (3\,j+4\,j+5\,j)[/tex]
[tex]\vec E = (-3-1-2)\,i + (3+4+5)\,j[/tex]
[tex]\vec E = -6\,i + 12\,j[/tex]
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants
Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
[tex]k = 0.903[/tex]
Explanation:
From the question we are told that
The time constant [tex]\tau = 3[/tex]
The potential across the capacitor can be mathematically represented as
[tex]V = V_o (1 - e^{- \tau})[/tex]
Where [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged
So at [tex]\tau = 3[/tex]
[tex]V = V_o (1 - e^{- 3})[/tex]
[tex]V = 0.950213 V_o[/tex]
Generally energy stored in a capacitor is mathematically represented as
[tex]E = \frac{1}{2 } * C * V ^2[/tex]
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at [tex]\tau = 3[/tex]
The energy stored can be evaluated at as
[tex]V^2 = (0.950213 V_o )^2[/tex]
[tex]V^2 = 0.903 V_o ^2[/tex]
Hence the fraction of the energy stored in an initially uncharged capacitor is
[tex]k = 0.903[/tex]
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?
Answer:
Explanation:
Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least and kinetic energy is maximum .
In this way , there is conservation of mechanical energy .
Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Answer:
C) experience a small induced magnetic moment when placed in an external magnetic field.
Explanation:
Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 3.62 g coins stacked over the 20.3 cm mark, the stick is found to balance at the 22.5 cm mark. What is the mass of the meter stick
Answer:
0.5792g
Explanation:
The computation of the mass of the meter stick is shown below:
Let us assume the following items
x1 = 50 cm;
m2 = m3 = 3.62 g;
x2 = x3 = 20.3 cm;
xcm = 22.5 cm
Based on the above assumption, now we need to apply the equation of center mass which is given below:
[tex]Xcm = \frac{m1x1 + m2x2 + m3x3}{m1 + m2 + m3} \\\\ 22.5 = \frac{m1\times 50 + 3.62 \times 20.3 + 3.62 \times 20.3}{m1 + 3.62 + 3.62}\\\\ 22.5m1 + 162.9 = 50m1 + 73.486 + 73.486[/tex]
27.5 m1 = 15.928
So, the m1 = 0.5792g