Answer:
If the rod is a conductor, the electrons are free to move within. So when the negative charge is brought towards the rod, the negatively charged electrons near the ball are repelled away towards the other end of the rod, leaving a net positive charge on the end of the rod near the ball. This causes the rod to be attracted towards the negatively charged ball and move closer.
Explanation:
Why is the earth Round
Answer:
because it has gravity in it . as it has more mass which can attract the things like we human are also attracted towards its centre and also mass is directly proportional to gravity which leads it to be round in shape.
hope u will get it..
Answer:
its round because of gravity
Explanation:
a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark
Answer:
-1748*10^N/C
Explanation:
See attached file
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the junction. How many electrons per second move past a point in wire 3?
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
[tex]i_1+i_3=i_2[/tex] (1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
Consider the momentum of a small ball during the projectile motion. Assume that there is no air friction. Is the momentum of the ball conserved
Answer:
Only the horizontal component of a projectile’s momentum is conserved. Where as The vertical component of the momentum is not conserved, because the net vertical force Fy–net is not zero
Explanation:
A centrifuge rotor is accelerated from rest to 20000 rpm in 30s a) what is its average angular acceleration b) through how many revolutions has the centrifuge rotor turned during it's acceleration period, assuming constant angular acceleration
Answer:
a. 70 rad/s²
b. 5000 rev
Explanation:
As we know,
[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]
then,
[tex]\omega=2100 \ rad/s[/tex]
a...
⇒ [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2100}{30}[/tex]
⇒ [tex]=70 \ rad/s^2[/tex]
b...
⇒ [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\times 70\times (30)^2[/tex]
[tex]=31500 \ rad[/tex]
[tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]
[tex]=5000 \ rev[/tex]
(a) The average angular acceleration will be 70 rad/s².
(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
What is angular acceleration?Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².
The given data in the problem is;
n is the revolution of centrifugal rotor = 20000 rpm
t is the time interval= the 30s
(α) is the Angular acceleration=?
n₁ is the revolution when the acceleration is constant =?
(a) The average angular acceleration will be 70 rad/s².
The value of the angular velocity is given by
[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]
The formula for angular acceleration is guven by;
[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]
Hence the average angular acceleration will be 70 rad/s².
(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.
[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]
As we know that the angular velocity is given by
[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]
The relation of angular velocity and revolution will be
[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]
Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
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Help yet again :) A hockey player is skating on the ice at 15km/h. He shoots the puck at 138 km/h according to a radar gun on the side of the ice. From the hockey player frame of reference how fast did he shoot the puck (in km/h)?
Answer:
speed of puck acc. to the radar gun = 138 km/h
speed of player = 15 km/h
since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,
speed of puck = speed of player + speed of puck acc. to player
138 = 15 + speed of puck acc. to player
speed of puck acc. to player = 138 -15
speed of puck acc. to player = 123 km/h
Brainly this answer if you think it deserves it
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
n electric motor rotating a workshop grinding wheel at 1.10 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 1.92 rad/s2. (a) How long does it take the grinding wheel to stop?
Answer:
Time taken to stop = 6 sec
Explanation:
Given:
Rotation of wheel = 1.10 × 10² rev/min
Final velocity (v) = 0
Angular acceleration (a) = - 1.92 rad/s²
Find:
Time taken to stop
Computation:
Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec
Initial velocity (u) = 11.52 rad / sec
We know that,
V = U +at
t = (v-u)a
t = (0 - 11.52) / (-1.92)
Time taken to stop = 6 sec
Three capacitors with capacitances:
C1=6.00μF,
C2=3.00μF,
C3=5.00μF.
The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 μC.
A. What is the charge Q1 on capacitor C1?
B. What is the charge on capacitor C3?
Express your answer in microcoulombs to three significant figures.
C. What is the applied voltage, Vab?
Express your answer in volts to three significant figures.
Answer:
A. 80.0microColoumbs
B.120.0 microcoloumbs
C.37.3v
Explanation:
See attached file
a. The charge Q1 on capacitor C1 should be 80.0 μC.
b. The charge on capacitor C3 should be 120 μC.
c. The applied voltage should be 37.5 v.
Calculation of the charge and voltage:a. We know that
v = q/c
Q1/c1 = Q2/C2
Q1 = (C1/C2) Q2
= (6.00/3.00) * 40
= 80.0
b.
Q3 = Q1 + Q2
= (80 + 40)
= 120
c.
The voltage should be
= 80/ 6+ 120/5
= 37.5 V
hence,
a. The charge Q1 on capacitor C1 should be 80.0 μC.
b. The charge on capacitor C3 should be 120 μC.
c. The applied voltage should be 37.5 v.
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a fly undergoes a displacement of - 5.80 while accelerating at -1.33 m/s^2 for 4.22 s. what was the initial velocity of the fly?
Answer:
[tex]v_i = 1.44\frac{m}{s}[/tex]
Explanation:
The computation of the initial velocity of the fly is shown below:-
But before that we need to do the following calculations
For 4.22 seconds
[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]
[tex]= -1.37\frac{m}{s}[/tex]
For uniform acceleration
[tex]\bar v = \frac{v_i +v_f}{2}[/tex]
[tex]= v_i + v_f[/tex]
[tex]= -2.74\frac{m}{s}[/tex]
With initial and final velocities
[tex]= -1.33\frac{m}{s^2}[/tex]
[tex]= \frac{v_i +v_f}{4.22s}[/tex]
[tex]= -v_i + v_f[/tex]
[tex]= -5.61\frac{m}{s}[/tex]
So, the initial velocity is
[tex]v_i = 1.44\frac{m}{s}[/tex]
We simply applied the above steps to reach at the final solution i.e initial velocity
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.
Answer:
Explanation:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4*10^6[2*2*10^-2/7*10^14]^1/2
=3*10^-2=3 cm
now we find the velocity f the electron strike the plate
v^2-(4*10^6)^2=2*7*10^14*2*10^-2
v^2=16*10^12+28*10^12
v^2=44*10^12
speed after hits =>V=6.6*10^6 m/s
Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.
Answer:
[tex]r_f=\frac{1}{6}r_i[/tex]
Explanation:
To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.
You use the following expression:
[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex] (1)
k: Coulomb's constant
qA: charge of A particle
qB: charge of B particle
When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:
[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex] (2)
To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:
[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]
The new separation of the charges is 1/6 times of the initial separation
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true?
A) The potential at the center of the sphere is zero.
B) The potential is lowest, but not zero, at the center of the sphere.
C) The potential at the center of the sphere is the same as the potential at the surface.
D) The potential at the center is the same as the potential at infinity.
E) The potential at the surface is higher than the potential at the center.
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
Cell membranes (the walled enclosure around a cell) are typically about d = 7.1 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 12.
Required:
a. What is the capacitance per square centimeter of such a cell wall?
b. In its normal resting state, a cell has a potential difference of 85mV across its membrane. What is the electric field inside this membrane?
Answer:
A. 1.24micro farad
B. 1.20x10^7V/m
Explanation:
See attached file
1.24micro farad is the capacitance per square centimeter of such a cell wall
In its normal resting state, a cell has a potential difference of 85mV across its membrane. the electric field inside this membrane is 1.20x10^7V/m.
what are the function of cell membrane ?The cell membrane is a selectively semi-permeable membrane surrounds the cytoplasm, the primary function of the membrane is to provide protection the integrity of the interior of the cell.
The cell membrane is called as multifaceted membrane which envelopes a cell's cytoplasm and help the cell to maintain the its shape.
Cell membrane is composed of Proteins and lipids where the mix or ratio of proteins and lipids can vary depending on the function of a specific cell.
Phospholipids are the crucial components of cell membranes which are arranged to form a lipid bilayer and certain substances can diffuse through the membrane to the cell's interior.
some cell organelles are surrounded by cell membranes, The nucleus and mitochondria have its own membrane.
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please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ball changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What’s the acceleration of the ball during the collision? Use . a= v-u/t
Answer:
the acceleration during the collision is: - 5 [tex]\frac{m}{s^2}[/tex]
Explanation:
Using the formula:
[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]
we get:
[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]
A parallel-plate capacitor C is charged up to a potential V0 with a charge of magnitude Q0 on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation.
A. What is the new capacitance in terms of C?
B. How much charge is now on the plates in terms of Q_0 ?
C. What is the potential difference across the plates in terms of V_0
Answer:
A. The new capacitance is half in term of C
B. The charge remains the same in terms of [tex]Q_{0}[/tex]
C. The potential difference is double in terms of [tex]V_{0}[/tex]
Explanation:
The battery with a voltage of [tex]V_{0}[/tex] is used to charge the plates, giving it a capacitance of C.
The charging process leaves a charge of magnitude [tex]Q_{0}[/tex] on the plate
The battery is disconnected (this will leave it with a constant charge [tex]Q_{0}[/tex])
the relationship between the charge, voltage and capacitance of the plate is
[tex]Q_{0}[/tex] = C[tex]V_{0}[/tex] ......... equ 1
A. The relationship between capacitance and the distance of the plate is given as
C = Aε/d ......... equ 2
where A is the area of the plate,
ε is the permeability of free space,
d is the distance between the plates
The area of the plate does not change, and permeability of free space is a constant. The combination of all these means that if the distance is doubled, then the capacitance will be halved. This is from equ 2 when the distance becomes 2d, then we have
C' = Aε/2d
==> C' = C/2
B. Since the battery is disconnected, and the capacitor is not discharged, the charge on the plate will remains the same as [tex]Q_{0}[/tex]. This is due to the conservation of charges.
C. Since the charge remains constant, and the capacitance is halved, then from equ 1, the new potential difference V will become double of the initial potential difference [tex]V_{0}[/tex]
==> V = 2[tex]V_{0}[/tex]
A merry-go-round is spinning at a rate of 3.0 revolutions per minute. Cora is sitting 1.0 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center.
1. What is the relationship between the rotational speed of the two children?
a. Cameron's rotational speed is double coral rotational speed.
b. Cameron's rotational speed is four times as much as coral rotational speed.
c. Cora rotational speed is double Cameron's rotational speed.
d. Cora rotational speed is the same as Cameron's rotational speed.
e. Cora rotational speed is four times as much as Cameron's rotational speed
2. What is the relationship between the tangential speed of the two children?
a. Cora tangential speed is four times as much as Cameron's rotational speed.
b. Camerons tangential speed is four times coral tangential speed.
c. Cora tangential speed is the same as Cameron's tangential speed.
d. Cora tangential speed is double Cameron's tangential speed.
e. Cameron's tangential speed is double coral tangential speed.
Answer:
Explanation:
1 )
angular velocity of merry go round = 2π n
= 2π x 3 / 60
ω = .1 x π radian / s
This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to
the centre of the merry go round .
So rotational speed of Cora = Rotational speed of Cameron
option ( d ) is correct .
2 )
Tangential speed v = ω R where R is the distance from the centre of merry go round .
Tangential speed of Cora = .1 x π x 1
= .314 m /s
Tangential speed of Cameron = .1 x π x 2
= .628 m /s .
So tangential speed of Cameron is twice that of Cora .
Option ( e ) is correct .
Answer:
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Explanation:
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the speed (in m/s) of the body at this instant?m/s(b)What is the mass (in kg) of the body at this instant?kg
Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
[tex]K = \frac{1}{2} mv^2[/tex]
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg
In a series RC circuit, the resistor voltage is 124 V and the capacitor voltage is 167 V. What is the total voltage
Answer:
208 V
Explanation:
resistor voltage = Vr = 124 V
capacitor voltage Vc = 167 V
the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor
total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]
==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V
a pendulum hanging from the ceiling of a train traveling with constant speed deviates 37 ° from the vertical when the train describes a curve of 60 m radius determines the train's rapids
Answer:
21 m/s
Explanation:
There are three forces on the pendulum:
Weight force mg pulling down,
Vertical tension component Tᵧ pulling up,
and horizontal tension component Tₓ pulling towards the center of the curve.
Sum of forces in the y direction:
∑F = ma
Tᵧ − mg = 0
T cos 37° = mg
T = mg / cos 37°
Sum of forces in the centripetal direction:
∑F = ma
Tₓ = mv²/r
T sin 37° = mv²/r
(mg / cos 37°) sin 37° = mv²/r
g tan 37° = v²/r
v = √(gr tan 37°)
v = √(9.8 m/s² × 60 m × tan 37°)
v = 21 m/s
A converging lens has a focal length of 14.0cm. For an object to the left of the lens, at distances of 18.0cm and 7.00cm, determine:
a. The image position
b. The magnification
c. Whether the image is real or virtual
d. Whether the image is upright or inverted
Answer:
Explanation:
A converging lens id also known as a convex lens. A convex lens has a positive focal length.
Using the lens formula to determine the image distance from the lens for each object distance.
1/f = 1/u + 1/v
f = focal length
u = object distance
v = image distance
For an object placed at distance of 18.0cm with focal length 14.0cm,
1/v = 1/f-1/u
1/v = 1/14 - 1/18
1/v = 9-7/126
1/v = 2/126
v = 126/2
v = 63cm
Since the image distance is positive for an object 18cm from the lens, the image formed by the object at this distance is a real and inverted image.
The magnification = v/u = 63/18 = 3.5
Similarly for an object placed at distance of 7.0cm with focal length 14.0cm,
v = uf/u-f
v = 7(14)/7-14
v= -14.0 cm
Since the image distance is negative for an object placed 7.0 cm from the lens, the image formed by the object at this distance is a virtual and upright image.
The magnification = v/u = 14/7 = 2
Note that the negative value is not taken into account when calculating magnification. The negative value only tells us the nature of the image formed.
A proton travels with a speed of 4.90 106 m/s at an angle of 56° with the direction of a magnetic field of magnitude 0.250 T in the positive x-direction. What is the magnitude of the magnetic force on the proton?
Answer:
F = 1510.21 x 10^-16N
Explanation:
F = Bqv Sin (angle)
F = 0.250 x 1.6 x 10^-19 x 4.9 x 10^6 x sin(56)
F = 1510.21 x 10^-16N
A clarinet behaves like a tube closed at one end. If its length is 1.0 m, and the velocity of sound is 344 m/s, what is its fundamental frequency (in Hz)
Answer:
Fundamental frequency = 1376Hz
Fundamental frequency of it is 86 Hz.
What is frequency?In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.
After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration.
Given parameters:
Length of the one end closed tube; L = 1 m.
Velocity of sound; v = 344 m/s.
We have to find, fundamental frequency of it: f₀ = ?
We know that frequency of one end closed tube is given by: f = (2n−1)v/4L where; n = 1, 2, 3, 4, .....
So, for n = 1; fundamental frequency is given by: f₀ = v/4L = 344/(4×1) Hz. = 86 Hz.
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A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will plot equipotential lines.
Answer:
E = - dV/dx
Explanation:
Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.
El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.
De los antes expuesto las dos magnitudes están relacionadas
E = - dV/dx
por lo cual el potenical es el gradiente del potencial eléctrico.
Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial
_____________ friction is the interlocking of surfaces due to irregularities on the surfaces preventing those surfaces from moving/sliding against each other. For surfaces moving/sliding on each other, ___________ friction overwhelms kinetic friction to that movement/sliding. Kinetic friction is alway larger than ____________ friction. Kinetic friction is alway equal to _________ friction.
Answer:
STATIC, STATIC
KINETIC friction is less than static friction
Explanation:
In this exercise you are asked to complete the sentences with the correct words.
STATIC friction prevents the relative movement of two surfaces in contact.
For moving surfaces the friction is STATIC is greater than the kinetic friction.
For the last two sentences I think they are misspelled, the correct thing is
KINETIC friction is less than static friction
Suppose the same magnitude force is applied at the same point as in the example, and the torque is found to have the same magnitude but in the opposite direction of the torque found there. What are the components of the force?
Answer:
-i - 7j
Explanation:
The computation of components of the force is shown below:-
torque T = r cross F
T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)
Now we will cross multiplying vectorilly
T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0
T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0
T = 28 × (-k) + 5 × (k) = -23k
net torque |T| = 23 N - m
direction >> negative k
Or Simply we can do by the below method
r × f
28 - 5 = 23 K
-i - 7j