Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?

The direction for each field vector is perpendicular to equipotential lines.

Take a snapshot of the simulation showing equipotential lines and paste to a word document.

Answer:

1. Transformer 3> Transformer 1 and 2 > Transformer 4

2. Transformer 2,3,5 > Transformer 1 > Transformer 4

Explanation:

From;

Vs/Vp = Ns/Np

Where;

Vp= voltage in primary coil

Vs= voltage in secondary coil

Ns= number of turns in secondary coil

Np= number of turns in primary coil.

Vs= Ns×Vp/Np

Vs= 480 ×2000/4000

Vs= 240 V

Vs= 480 ×1000/2000

Vs=240 V

Vs= 120 × 2000/500

Vs= 480 V

Vs= 240 × 500/1000

Vs= 120 V

2. Ns/Np= Ip/Is

Is= Np×Ip/Ns

Is= 1000 × 100/2000

Is= 50 A

Is= 2000 × 100/1000

Is= 200 A

Is= 1000 × 100/500

Is= 200 A

Is= 500 × 100/2000

Is= 25 A

Is= 4000 × 100/2000

Is= 200 A

Answer:

Knowledge of vectors is important because many quantities used in physics are vectors. If you try to add together vector quantities without taking into account their direction you'll get results that are incorrect.

Explanation:

An example of the importance of vector addition could be the following:

Two cars are involved in a collision. At the time of the collision car A was travelling at 40 mph, car B was travelling at 60 mph. Until I tell you in which directions the cars were travelling you don't know how serious the collision was.

The cars could have been travelling in the same direction, in which case car B crashed into the back of car A, and the relative velocity between them was 20 mph. Or the cars could have been travelling in opposite directions, in which case it was a head on collision with a relative velocity between the cars of 100 mph!

Answer:

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

Hence, the resistance of this bulb is 121 ohms

Answer:

10%

Explanation:

Efficiency = work done / energy used

e = (10 m × 100 N) / (10,000 J)

e = 0.1

The efficiency is 0.1, or 10%.

Answer:

380 kHz

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

where v is the speed of a wave

λ is the wavelength of the wave

f is the frequency of the wave

f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ 380 kHz

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

Answer:

they must be affordable because they have to pay for it or they wont get the stuff they are bying.

Explanation:

need a brainliest please.

Answer: B, they must be affordable.

Explanation:

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

Answer:

Explanation:

Let B= bead

Q = rod

the electric field at the glass bead pocation is

(Gauss theorem)

E = Q / (2 π d L εo)

the force is

F = q E = q Q / (2 π d L εo)

then

Q = 2 π d L εo F / q

Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC

Answer:

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Explanation:

A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.

Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.

So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

[tex]m1U1+ m2U2= m1V1+m2V2[/tex]

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

[tex]m1U1+ m2U2= V(m1+m2)[/tex]

Answer:

testable

Explanation:

high heat can cause browning and or welting

Answer:

A

Explanation:

If it is tested, it will really changed when the plant into low temperature

Answer:

There is no induced current on the coil.

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

Answer:

The projectile strikes the tower at a height of 354.824 meters.

Explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]

Vertical motion

[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.

[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]

If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:

[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]

[tex]t \approx 7.071\,s[/tex]

Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])

[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]

[tex]y \approx 354.824\,m[/tex]

The projectile strikes the tower at a height of 354.824 meters.

Answer:

The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]

Explanation:

From the question we are told that

   The  diameter is  [tex]d = 60 \ mm = 0.06 \ m[/tex]

    The  head is  [tex]h = 6 \ m[/tex]

     The  coefficient of contraction is  [tex]Cc = 0.68[/tex]

     The  coefficient of  velocity is  [tex]Cv = 0.92[/tex]

The radius is mathematically evaluated as

         [tex]r = \frac{d}{2}[/tex]

substituting values

        [tex]r = \frac{ 0.06 }{2}[/tex]

        [tex]r = 0.03 \ m[/tex]

The  area is mathematically represented as

      [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.03)^2[/tex]

      [tex]A = 0.00283 \ m^2[/tex]

 The  discharge rate is mathematically represented as

        [tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]

substituting values

       [tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]

       [tex]Q = 0.0192 \ m^3 /s[/tex]

Answer:

C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.

Explanation:

The complete question is

Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.

Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.

Answer:

Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."

Explanation:

My options were:

A)  Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.

B)  Resonance occurs as a result of sympathetic vibrations.

C)  A non-vibrating object can begin to vibrate as a result of forced vibrations.

D)  Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.

A is correct

.

Answer::

   [tex]\lambda = 634 nm[/tex]

  [tex]f = 4.73 *10^{14} \ Hz[/tex]

Explanation:

From the question we are told that

      The  distance of separation is  [tex]d = 0.047 \ mm = 0.047 *10^{-3} \ m[/tex]

       The  distance of the screen is  [tex]D = 6.60 \ m[/tex]

      The  width of the fringe is [tex]y = 8.9 \ cm = 0.089 \ m[/tex]

Generally the width of the width of the fringes is mathematically represented as

          [tex]y = \frac{\lambda * D }{d }[/tex]

=>       [tex]\lambda = \frac{y * d }{D }[/tex]

=>      [tex]\lambda = \frac{ 0.089 * (0.047 *10^{-3}) }{6.60 }[/tex]

=>    [tex]\lambda = 634 *10^{-9}[/tex]

=>  [tex]\lambda = 634 nm[/tex]

Generally the speed of light is mathematically represented as

         [tex]c = f * \lambda[/tex]

=>       [tex]f= \frac{ c}{\lambda }[/tex]

=>         [tex]f= \frac{ 3.0 *10^{8}}{634 *10^{-9}}[/tex]

=>     [tex]f = 4.73 *10^{14} \ Hz[/tex]

Answer:

Mass of the car is independent of gravity

Explanation:

Here, we want to state the reason why even though we have the acceleration due to gravity absent on the moon, it is still difficult to accelerate a car on a level horizontal level on the moon.

The answer to this is that the mass of the car that we want to accelerate is independent of gravity.

Had it been that gravity has an effect on the mass of the said car, then we might conclude that it will not be difficult to accelerate the car on a horizontal surface on the moon.

But due to the fact that gravity has no effect on the mass of the car to be accelerated, then the problem we have on earth with accelerating the car is the same problem we will have on the moon if we try to accelerate the car on a horizontal level surface.

Answer:

Explanation:

Christian Huygens

Light Is a Wave!

Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.

Answer:

B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction

Explanation:

In an electromagnetic wave the electric and magnetic fields are in phase

         c = E / B

         B = E / c

let's calculate

          B = 4.00 / 3 10⁸

          B = 1.33 10⁻⁸ T

To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.

 If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction

Answer:

The hoop

Explanation:

Because it has a smaller calculated inertia of 2/3mr² compares to the disc

Answer: μ = 0.8885

Explanation: Force due to friction is calculated as: [tex]F_{f} = \mu.N[/tex]

At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.

Power associated with work done by friction is [tex]P=F_{f}.x[/tex]. The variable x is displacement the object "spent its energy".

Power associated with work done by gravitational force is P = mghcosθ, where h is height.

The decline forms with horizontal plane a triangle as draw in the picture.

To determine force due to friction:

[tex]F_{f}.x=mghcos(\theta)[/tex]

[tex]F_{f}=\frac{mghcos(\theta)}{x}[/tex]

Replacing force:

[tex]\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)[/tex]

[tex]\mu=\frac{h}{x}[/tex]

Calculating h using trigonometric relations:

[tex]sin(32.51) = \frac{h}{1}[/tex]

h = sin(32.51)

Coefficient of Kinetic friction is

[tex]\mu=\frac{sin(32.51)}{1}[/tex]

μ = 0.8885

For these conditions, coefficient of kinetic friction is μ = 0.8885.

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

Answer:

t = 23.6 min

Explanation:

First we need to find the mass of air in the room:

m = ρV

where,

m = mass of air in the room = ?

ρ = density of air at room temperature = 1.2041 kg/m³

V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³

Therefore,

m = (1.2041 kg/m³)(3247.2 m³)

m = 3909.95 kg

Now, we find the amount of energy consumed to heat the room:

E = m C ΔT

where,

E = Energy consumed = ?

C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C

ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C

Therefore,

E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)

E = 29324.62 KJ

Now, the time period can be calculated as:

P = E/t

t = E/P

where,

t = Time needed = ?

P = Power of heater = 20.7 KW

Therefore,

t = 29324.62 KJ/20.7 KW

t = (1416.65 s)(1 min/60 s)

t = 23.6 min

Answer:

Wavelength is 471 nm

Explanation:

Given that,

Lines per unit length of diffraction grating is 500 lines/mm.

The third maxima from the central maxima (m=3) is at an angle of 45°

We need to find the color of laser light shines through a diffraction grating.

The condition for maxima is :

[tex]d\sin\theta=m\lambda[/tex]

d = 1/N, N = number of lines per mm

[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]

Answer:

To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D

Answer:

292 nm

Explanation:

The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J

at maximum wavelength, the energy of the photons is equal to its work function

Energy of the electron = hf

but hf = hc/λ

where h is the planck's constant = 6.63 × 10-34 m^2 kg/s

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength that this occurs, which is the maximum wavelength

Equating, we have

hc/λ =  ∅

substituting, we have

(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19

(1.989 x 10^-25)/(6.81 x 10^-19) = λ

λ = 292.07 x 10^-9 = 292 nm

Answer:

a) m = 0.626 kg , b) T = 2.09 s , c)   a = 1.0544 m / s²

Explanation:

In a spring mass system the equation of motion is

        x = A cos (wt + Ф)

with      w = √(k / m)

a) velocity is defined by

        v = dx / dt

        v = - A w sin (wt + Ф)          (1)

give us that the speed is

        v = 26.1 m / s

for the point

        x = a / 2

the range of motion is a = 11.0 cm

       x = 11.0 / 2

       x = 5.5 cm

Let's find the time it takes to get to this distance

       wt + Ф = cos⁻¹ (x / A)

       wt + Ф = cos 0.5

        wt + Ф = 0.877

In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f

       Ф = 0

       wt = 0.877

       t = 0.877 / w

we substitute in equation 1

       26.1 = -11.0 w sin (w 0.877 / w)

        w = 26.1 / (11 sin 0.877))

        w = 3.096 rad / s

from the angular velocity equation

       w² = k / m

       m = k / w²

       m = 6 / 3,096²

       m = 0.626 kg

b) angular velocity and frequency are related

       w = 2π f

frequency and period are related

        f = 1 / T

we substitute

        w = 2π / T

        T = 2π / w

        T = 2π / 3,096

        T = 2.09 s

c) maximum acceleration

 the acceleration of defined by

        a = dv / dt

        a = - Aw² cos (wt)

the acceleration is maximum when the cosine is ±1

         a = A w²

          a = 11  3,096²

        a = 105.44 cm / s²

we reduce to m / s

        a = 1.0544 m / s²

Answer:

[tex]r=4.24\times 10^{11}\ m[/tex]

Explanation:

Given that,

Orbital time period, T = 3.75 earth years

Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]

We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.

[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]

r is the radius of the exoplanet's orbit.

Solving for r we get :

[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]

We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]

So,

[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]

So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].