Answer:
It will take. the same distance up as before, but take a longer time
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed:_____
a. into the page.
b. toward the left
c. toward the right
d. toward the bottom of the page.
e. toward the top of the page.
f. out of the page.
Answer: F
Out of the page.
Explanation:
For an electron with a charge of -e, the magnitude of the force on it is F = BeV
Where
F = force on the electron
e = charge ( electrons )
V = velocity
B = magnetic field
F is the force acting on all the electrons in a wire which gives rise to the F = BIL
Where
I = current
L = length of the wire
The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.
A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend
Answer:
Explanation:
The e.m.f induced in the coil depend on the following :
(a) No. of turns in the coil
(b) Cross-sectional Area of the coil
(c) Magnitude of Magnetic field
(d) Angular velocity of the coil
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g
Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
for an answer to be complete,the units needs to be specified.why
Explanation:
unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998.0 rad/s while it is being read, and then is allowed to come to rest over 0.502 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
1988.05 rad/s^2
Explanation:
The angular speed of the optical disk ω = 998.0 rad/s
the time taken to come to rest t = 0.502 s
The magnitude of the average angular acceleration ∝ = ω/t
∝ = 998.0/0.502 = 1988.05 rad/s^2
Which of the following frequencies could NOT be present as a standing wave in a 2m long organ pipe open at both ends? The fundamental frequency is 85 Hz.
Answer:
382Hz
Explanation:
The question lacks the required option. Find the complete question in the attachment.
The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L
Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...
Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;
First overtone f1 = 2fo = 2(85) = 170Hz
Second overtone f2 = 3fo = 3(85) = 255Hz
Third overtone = 4fo = 4(85) = 340Hz
Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz
If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away (in m) is the planet Venus if the echo time is 900 s? m (b) What is the echo time (in µs) for a car 80.0 m from a Highway Patrol radar unit? µs (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 11.5 m? ns
Answer:
a) 1.35 x 10^11 m
b) 0.53 µs
c) 8 ns
Explanation:
Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s
a) for 900 s,
The distance for a round trip = v x t
==> (3 x 10^8) x 900 = 2.7 x 10^11 m
The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m
b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of
d = 2 x 80 = 160 m
the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs
c) accuracy in distance Δd = 11.5 m
Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns
An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.
Required:
a. What is the angular frequency of the circuit when R = 0?
b. What value must R have to give a decrease in angular frequency of 10.0 % compared to the value calculated in Part a.
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.
b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.
Given the following data:
Inductance, L = 0.450 HenryCapacitance, C = [tex]2.50\times 10^{-5}[/tex] Faradsa. To determine the angular frequency of the circuit when R = 0 Ohms:
Mathematically, the angular frequency of a LC circuit is given by the formula:
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]
b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:
The mathematical expression is given as follows:
[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]
[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]
R = 116.96 Ohms.
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A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer?
O The nail exerts a much smaller force on the hammer in the opposite direction
O The nail exerts a much smaller force on the hammer in the same direction.
The nail exerts an equal force on the hammer in the same direction.
O The nail exerts an equal force on the hammer in the opposite direction.
Answer:
reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.
This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.
The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of the hammer is much greater, therefore the pressure is small.
When reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?
Answer:
t = t₀ / 2
Explanation:
In this exercise we must use Newton's second law
F = m a
a = F / m
now we can use kinematics
as in object part of rest (v₀ = 0)
v =a t₀
t₀ = v / a
these results are with the first experiment
now repeat the experiment, but F = 2F₀
a = 2F₀ / m = 2 a₀
v = 2 a₀ t
t = v / 2a₀
t = t₀ / 2
The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].
Given the following data:
Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = VTo find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]F = \frac{M(V-U)}{t}[/tex]
Where:
F is the force.V is the final velocity.U is the initial velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]
When the experiment is repeated, the magnitude of the force is doubled:
[tex]F = 2F[/tex]
Now, we can find the time interval that is required to reach the same final speed (V):
[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]
Substituting the value of F, we have:
[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]
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Find the angle in degrees for the third-order maximum for 577 nm wavelength yellow light falling on a diffraction grating having 1,420 lines per centimeter.
Answer:
θ = 0.14°
Explanation:
Here we will use the grating equation. The grating equation is as follows:
mλ = d Sin θ
where,
θ = angle = ?
m = order number = 3
λ = wavelength of light = 577 nm = 5.77 x 10⁻⁷ m
d = spacing between slits = 1/(1420 lines/cm) = 7.042 x 10⁻⁴ m
Therefore, using these values, we get:
(3)(5.77 x 10⁻⁷ m) = (7.042 x 10⁻⁴ m)Sin θ
Sin θ = (3)(5.77 x 10⁻⁷ m)/(7.042 x 10⁻⁴ m)
Sin θ = 2.46 x 10⁻³
θ = Sin⁻¹(2.46 x 10⁻³)
θ = 0.14°
The magnitude of the magnetic field at point P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at P? (c = 3.0 × 108 m/s)
Answer:
The electric field is [tex]E = 636 \ V/m[/tex]
Explanation:
From the question we are told that
The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]
The value for speed of light is [tex]c = 3.0 *10^8 \ m/s[/tex]
Generally the magnitude of the electric field at point P is
[tex]E = B * c[/tex]
substituting values
[tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]
[tex]E = 636 \ V/m[/tex]
The magnitude of electric field for the wave at point P is 636 V/m.
Given data:
The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].
The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].
The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,
[tex]E = B \times c[/tex]
here,
E is the strength of electric field.
Solving as,
[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]
Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.
Learn more about the electric field here:
https://brainly.com/question/15800304
If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g
Answer:
the voltage drop across this same diode will be 760 mV
Explanation:
Given that:
Temperature T = 300°K
current [tex]I_1[/tex] = 100 μA
current [tex]I_2[/tex] = 1 mA
forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V
To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.
Using the formula:
[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
where;
[tex]V_r[/tex] = 0.7
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]
[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
Suppose n = 1
[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]
Then;
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]
[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]
[tex]V_r'=29.37 \times nV_T[/tex]
[tex]V_r'=29.37 \times 25.86[/tex]
[tex]V_r'=759.5 \ mV[/tex]
[tex]Vr' \simeq[/tex] 760 mV
Thus, the voltage drop across this same diode will be 760 mV
Which one of the following lists gives the correct order of the electromagnetic spectrum from low to high frequencies?
A) radio waves, infrared, microwaves, ultraviolet, visible, x-rays, gamma rays
B) radio waves, ultraviolet, x-rays, microwaves, infrared, visible, gamma rays
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
D) radio waves, microwaves, visible, x-rays, infrared, ultraviolet, gamma rays
E) radio waves, infrared, x-rays, microwaves, ultraviolet, visible, gamma rays
Answer:
C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays
Explanation:
radio waves have lowest energy , lowest frequency and highest wavelength
gamma rays have highest energy , highest frequency and least wavelength
Answer: C
Explanation:
Find the minimum thickness (in nm) of a soap bubble that appears green when illuminated by white light perpendicular to its surface. Take the wavelength to be 549 nm, and assume the same index of refraction as water (nw
Answer:
103nm
Explanation:
Pls see attached file
A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up from chest height to full arm extension, a distance of 0.53 m .
Part A: How much work does the weightlifter do to lift the barbell one time?
Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body?
Part C: How many 500 Calorie donuts can she eat a day to supply that energy?
Answer:
A) Workdone = 223.57 N-m
B) 22357 J of energy
C) Number of donuts = 10.7 donuts
Explanation:
A) The work done is calculated from the formula;. Work done = Force × Distance
We are given;
Mass; m = 43 kg
Distance = 0.53 m
Force(weight) = mg = 43 × 9.81
Thus;
Work done = 43 × 9.81 × 0.53
Workdone = 223.57 N-m
B) We are told she does 23 repetitions a day.
Thus, we assume 23% efficiency.
So, Work = Energy
Thus;
At 100% efficiency;
Energy = (223.57/100%) × 23 repetitions = 5142.11 J
Now, since she is only 23% efficient, she will expend; 5142.11/0.23 J = 22357 J of energy to do 5390 J of work.
C) from conversions; 4.18 J = 1 calorie
Thus;
22357 J ÷ 4.18 J/cal = 5348.565 calories
We how many 500 calorie donuts she can eat in a day to supply that energy.
Thus;
Number of donuts = 5348.565 cal ÷ 500 cal /donut
Number of donuts = 10.7 donuts
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .
Requried:
What is the magnitude of the charge (in nC) on each bead?
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is the angle of refraction in degrees?
Answer:
θ = 10.28º
Explanation:
To find the angle of refraction use the equation of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for incident light and index 2 is for refracted light.
sin θ₂ = n₁ / n₂ sin θ
let's calculate
sin = 1 / 1.3 sin 0.23
sin = 0.175
θ= 0.17528 rad
let's reduce to degrees
θ = 0.17528 rad (180ª / pi rad)
θ = 10.28º
g A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75 C/s . Part A How many electrons are delivered to the
Answer:
Approximately 5 x 10^18 electrons are delivered to the smart phone charger.
Explanation:
The electric current in a circuit is the flow of charges through a circuit with time.
The charges through the circuit are due to the electrons that flow through the circuit.
An individual electrons has a charge of -1.60 x 10^-19 C on it.
If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.
==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.
If a train travels at a constant 18.0 m/s, how far would it move in one hour? In 1.00 minute? In 1.00 second?
Explanation:
Distance = speed × time
d = (18.0 m/s) (1 hr × 3600 s/hr)
d = 64,800 m
d = (18.0 m/s) (1 min × 60 s/min)
d = 1080 m
d = (18.0 m/s) (1 s)
d = 18.0 m
There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,
1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N
Explanation:
Pressure is the same for both plungers.
P = P
F / A = F / A
F / (¼ π d²) = F / (¼ π d²)
F / d² = F / d²
5 N / (0.05 m)² = F / (1 m)²
F = 2000 N
None of the options are correct.
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the collision. What is the wavelength of the scattered photon?
0.6764*10^-10m
Explanation:
Using E= hc/wavelength
(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV
So subtract the calculated energy from the given energy of scattered photons
9.11-0.75=18.36 keV
To find wavelength
Wavelength= hc/ E
[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m
A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?
Answer:
The new time period is [tex]T_2 = 3.8 \ s[/tex]
Explanation:
From the question we are told that
The period of oscillation is [tex]T = 5 \ s[/tex]
The new length is [tex]l_2 = 0.76 \ m[/tex]
Let assume the original length was [tex]l_1 = 1 m[/tex]
Generally the time period is mathematically represented as
[tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]
Now I is the moment of inertia of the stick which is mathematically represented as
[tex]I = \frac{m * l^2 }{12 }[/tex]
So
[tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]
Looking at the above equation we see that
[tex]T \ \ \ \alpha \ \ \ l[/tex]
=> [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]
=> [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]
=> [tex]T_2 = 3.8 \ s[/tex]
Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm
Answer:
R = 0.992 Ω
Explanation:
En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.
Matemáticamente,
Resistencia = (resistividad * longitud) / Área De la pregunta;
Resistividad = 3,5 * 10 ^ -5 Ωm
longitud = 170 cm = 1,7 m
Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2
Conectando estos valores a la ecuación anterior, tenemos;
Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =
(3.5 * 1.7) / 6 = 0.992 Ω
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d 1.53 cm and a plate area of A = 25.0 cm2. when the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
(a) Calculate the charge on the plates in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before Qi = _____
after Qf = ______
(b) Determine the capacitance (in F) and potential difference (in V) after immersion
(c) Determine the change in energy (in n]) of the capacitor Δυ = nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 275 V potential difference
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
Determine the capacitance (in F) and potential difference (in V) after immersion
Determine the change in energy (in nJ) of the capacitor AU nJ
Answer:
a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,
c) U = 54.7 nJ , d) ΔU = 54 nJ,
Explanation:
a) The capacity of a capacitor is defined
C = Q / V
Q = C V
can also be calculated using geometry consideration
C = e or A / d
we reduce to the SI system
A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²
d = 1.53 cm = 1.53 10⁻² m
we substitute
Q = eo A / d V
Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275
Q = 3.9757 10⁻¹⁰ C
let's reduce to pC
Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)
Q = 397.57 pC
when the capacitor is introduced into the water the dielectric constant is different
Q = k Q₀
Q = 80 397.57
Q = 3.18 104 pC
b) Find capacitance and voltage after submerged in water
C = k C₀
C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²
C = 1.157 10⁻¹⁰ F
V = Vo / k
V = 275/80
V = 3.4375 V
c) The stored energy is
U = ½ C V²
U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²
U = 5.47 10⁻⁸ J
let's reduce to nJ
109 nJ = 1 J
U = 54.7 nJ
d) energy after submerging
U = ½ (kCo) (Vo / k) 2
U = ½ Co Vo2 / k
U = U₀ / k
U = 54.7 / 80 nJ
U = 0.68375 nJ
the energy change is
ΔU = U₀ -U
ΔU = 54.7 - 0.687375
(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.
(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.
(c) The change in energy of the capacitor is 54.02 nJ.
Charge on the plate before immersionThe charge on the plate is calculated as follows;
[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]
Charge on the plate after immersion[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]
Capacitance and potential difference after immersion[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]
[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]
Change in energy of the capacitorThe initial energy of the capacitor is calculated as follows;
[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]
The final energy of the capacitor is calculated as follows;
[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]
Change in energy is calculated as follows;
[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]
Learn more about energy stored in a capacitor here: https://brainly.com/question/13578522
You are performing an experiment that requires the highest-possible magnetic energy density in the interior of a very long current-carrying solenoid. Which of the following adjustments increases the energy density?a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed. b. Increasing the number of turns per unit length on the solenold. c. Increasing the cross-sectional area of the solenoid. d. None of these. e. Increasing the current in the solenoid.
Answer:
The correct choice is B & E.
Explanation:
A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.
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To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.0 m/s. How much time does it take for the glove to return to the pitcher
Answer:
The glove takes 1.02s to return to the pitchers hand.
Explanation:
Given;
initial velocity the pitcher's glove, u = 5 m/s
Apply kinematic equation
s = ut - ¹/₂gt²
where;
g is acceleration due to gravity = 9.8 m/s²
t is the time takes the glove to return to the pitchers hand
s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)
0 = ut - ¹/₂gt²
ut = ¹/₂gt²
u = ¹/₂gt
gt = 2u
t = (2u) / g
t = (2 x 5) / 9.8
t = 1.02 s
Therefore, the glove takes 1.02s to return to the pitchers hand.
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
