Distinguish between physical and chemical changes. Include examples in your explanations.

Answer:

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

Explanation:

The transformer turns ratio is given by the following equation:

V₁/V₂ = N₁/N₂

where,

V₁ = Voltage of outlet = 120 V

V₂ = Device Voltage = 3 V

N₁ = No. of turns on outlet side = ?

N₂ = No. of turns on side of device = 45

Therefore,

120 V/3 V = N₁/45

N₁ = (40)(45)

N₁ = 1800 turns

So, the side of the transformer that plugs into the outlet has 1800 turns.

Answer:

A. 0.0374C

B. 0.012F

C. 18 ohms

Explanation:

See attached file

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".

The region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.

According to the question,

Current, I = 2.4 A

Radius, R = 5 cm or,

                = 0.05 m

Constant, μ₀ = 4π × 10⁻⁷ H/m

We know the relation,

The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]

By substituting the values in the above relation, we get

                                    = [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]

                                    = 9.6 × 10⁻⁴ T

Thus the above answer is appropriate.

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The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Answer:

Hey there!

PE=mgh, so as height decreases, so does the potential energy.

KE=mv^2, so as velocity increases, kinetic energy increases.

Thus, the correct answer would be Decreases, Increases.

Let me know if this helps :)

Answer:

energy carried by the current is given by the pointyng vector

Explanation:

The current is defined by

       i = dQ / dt

this is the number of charges per unit area over time.

The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.

But the energy carried by the current is given by the pointyng vector of the electromagnetic wave

            S = 1 / μ₀ EX B

It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]

Put the value into the formula

[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]

[tex]\lambda_{avg}=550\ nm[/tex]

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

[tex]w=\dfrac{2.44\lambda L}{D}[/tex]

[tex]D=\dfrac{2.44\lambda L}{w}[/tex]

Put the value into the formula

[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]

[tex]D=0.537\ mm[/tex]

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

Answer:

It will be half that if the first wave

Explanation:

Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it

Answer:

The magnification is  [tex]m = 12[/tex]

Explanation:

From the question  we are told that

   The object distance is [tex]u = 36.2 \ cm[/tex]

     The focal length is  [tex]v = 39.5 \ cm[/tex]

From the lens equation we have that

         [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]

=>     [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]

substituting values

       [tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]

       [tex]\frac{1}{v} = -0.0023[/tex]

=>   [tex]v = \frac{1}{0.0023}[/tex]

=>   [tex]v =-433.3 \ cm[/tex]

The magnification is mathematically represented as

         [tex]m =- \frac{v}{u}[/tex]

substituting values

        [tex]m =- \frac{-433.3}{36.2}[/tex]

         [tex]m = 12[/tex]

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

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]

Answer:

Explanation:

Answer:

quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.

Explanation:

Answer:

Option (A) : Nuclear Fusion and Beta Decay (electron emission)

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

Answer:

variable

Explanation:

Answer:

As in explanation.

Explanation:

A) Centre of Curvature: This is defined as the point in the center of the sphere from which the mirror was sliced. It is represented by the letter "C"

B) Vertex: It is defined as the point on the mirror's surface where the principal axis meets the mirror. It is represented by the letter A.

C) Focal Point: This is defined as the Midway point between the vertex and the center of curvature. It is represented by the letter "F"

D) Radius of Curvature: This is defined as the distance from the vertex to the center of curvature. It is represented by the letter "R"

E) Focal Length: This is defined as the distance from the mirror to the focal point. It's represented by the letter "f"

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]

where [tex]F/l[/tex] is the force per unit length on the wires

[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A

[tex]I_{1}[/tex] = current on the first wire = 0.520 A

[tex]I_{2}[/tex] = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] =  [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]

4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]

[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]

and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:

[tex]E=\frac{\Delta\,V}{d}[/tex]

then :

[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]

We want to find when the particle reaches velocity zero via kinematics:

[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]

We replace this time (t) in the kinematic equation for the particle displacement:

[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]

Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:

[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]

we get:

[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

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:

To 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:

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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Answer:

there are over 100 billion stars in our galaxy.

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

Answer:

The tension in the cord provides by centripetal force

T = Fc

= mv^2/r

= 2kg ( 6)^2/2

=36 N

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.

Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.

From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)

Where:

[tex]W_{T}[/tex] - Work done by tension, in joules.

[tex]m[/tex] - Mass of the sled-victim system, in kilograms.

[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.

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

[tex]s[/tex] - Travelled distance, in meters.

[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.

[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.

If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]

[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]

[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]

[tex]W_{T} = 1662.544\,J[/tex]

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

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