An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.

Required:
Find the self-induced emf in this inductor.

Answer:

Explanation:

Let the critical angle be C .

sinC = 1 / μ where μ is index of refraction .

sinC = 1 /1.2

= .833

C = 56°

Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )

sin i / sinr = 1.2 , i is angle of incidence

sini = 1.2 x sinr = 1.2 x sin 34 = .67

i = 42°.  

Answer:

Yes, it is possible for a radioactive nucleus to decay two times and end up as the same element as the original

Explanation:

Radioactive decay of nucleus is the process by which an unstable atomic nucleus loses energy by radiation. The energy loss may be from an excited nucleus, which may be emitted as a gamma ray in a process called gamma decay or due to a mass deficit whenever the nucleus decays according to the energy-mass relationship. Whenever a radioactive nucleus decays, new nucleus may be formed in a "transmutation decay", usually into an element of a lower atomic mass than the mother element. Another type of radioactive decay results in products that vary, appearing as two or more "fragments" of the original nucleus with a range of possible masses. This decay is called "spontaneous fission" and it usually happens when a large unstable nucleus decays into two (or occasionally three) smaller daughter nuclei, with an emission of gamma rays, neutrons, or other particles.

Explanation:

It is given that,

A nearsighted woman has a far point of 180 cm.

The object distance for nearsightedness is infinity, u = ∞

The image distance is, v = -180 cm

Using lens formula,

[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-180)}-\dfrac{1}{\infty}\\\\f=-180\ cm\\\\f=-1.8\ m[/tex]

For nearsightedness, a diverging lens is used.

Power of the lens,

P = 1/f

[tex]P=\dfrac{1}{-1.8}\\\\P=-0.56\ D[/tex]

So, the power of the lens is -0.56 D.

The power of lens to be prescribed is 3.5 D.

A person that is far sighted can see far objects clearly but not nearby objects. The farpoint of a normal eye is infinity. Far sightedness is corrected by the use of a convex lens. The near point of the normal eye is 25 cm.

We have that;

u = 25 cm

v = -180 cm

1/f = 1/u - 1/v

1/f = 1/25 - 1/180

1/f = 0.04 - 0.0056

f = 29 cm

Power of lens = 100/f = 100/29 = 3.5 D

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Answer:

The correct answer is B

Explanation:

The lunar month is the time it takes for the moon to go around the Earth, for this we use Newton's second law where the force is the universal force of attraction

         F = ma

the universal attractive force is

           F = G M_earth m_moon / r²

Accelerations centripetal acceleration

           a = w² r

the angular velocity is for the movement is

         w = 2π / T

Where T is the period of revolution; let's substitute

         G M_eath m_moon / r² = m_moon (2π /T)² r

          G M_earth = 4π² / T² r³

           T² = (4π² / G M_eart) r³

Let's analyze this equation we see that it does not depend on the mass of the Luma, therefore the period is the same

The correct answer is B

Answer:

edqr3fr =wdmmwfmnl[=kvlfkgvb

Explanation:bcz idk

Me BeLIves iT iS ZeRo So A

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

Answer:

0.514 m/s

Explanation:

The knot is a unit of nautical speed used in maritime navigation.

1 knot is equal to one nautical mile per hour,

1 nautical mile per hour = 1.852 km/h

The basic SI system of units of speed is in 'm/s'

1.852 km/h = (1.852 x 1000)/3600 = 0.514 m/s

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

Answer: B. A magnet pulls a nail towards it

Explanation: Out of the list of answers, this is the only answer where a force is acting on an object. the act of the magnet pulling on the nail is the force, and the nail is the object.

Complete Question

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s.

Part A What is the electric field strength inside the solenoid at a point on the axis?

Part B

What is the electric field strength inside the solenoid at a point 1.50 cm from the axis?

Answer:

Part A

    [tex]E = 0 \ V/m[/tex]

Part  B

  [tex]E_{15} = 0.0345 \ V/m[/tex]

Explanation:

From the question we are told that

    The diameter of the solenoid is  [tex]d = 5.0 \ cm = 0.05 \ m[/tex]

    The magnetic field is  [tex]B = 2.0 \ T[/tex]

     The  rate of the change of the magnetic field is  [tex]\frac{dB}{dt} = 4.60 \ T/s[/tex]

The radius of the solenoid is mathematically represented as

         [tex]R = \frac{ d}{2}[/tex]

substituting values

        [tex]R = \frac{ 5.0 *10^{-2}}{2} = 0.025 \ m[/tex]

Generally the of the solenoid is mathematically  represented as

        [tex]E = \frac{ r}{2} * |\frac{dB}{dt} |[/tex]

Now at the point on axis is r = 0 given that the axis is the origin so

      [tex]E = \frac{ 0}{2} * |\frac{dB}{dt} |[/tex]

      [tex]E = 0 \ V/m[/tex]

Now  the electric field strength inside the solenoid at a point 1.50cm from the axis is  mathematically represented as

        [tex]E_{15} = \frac{ 15*10^{-2 }}{2} * |4.60 |[/tex]

       [tex]E_{15} = 0.0345 \ V/m[/tex]

Answer:

0.750 J

Explanation:

69% of the elastic energy is converted to kinetic energy.

0.69 EE = KE

0.69 EE = ½ mv²

EE = mv² / 1.38

EE = (0.0342 kg) (5.50 m/s)² / 1.38

EE = 0.750 J

Answer:

m = 469.51

Explanation:

Given that,

The diameter of a refracting telescope is 1.48 m

Focal length of the objective lens is 15.4 m

Focal length of an eyepiece is 3.28 cm

We need to find the angular magnification of the telescope. The ratio of focal length of objective lens to the focal length of the eye piece is called angular magnification of the telescope. So,

[tex]m=\dfrac{f_o}{f_e}\\\\m=\dfrac{15.4}{3.28\times 10^{-2}}\\\\m=469.51[/tex]

So, the angular magnification of the telescope is 469.51.

Answer:

BS33

Explanation:

its a cable length meter used for measuring length of all kinds of wires.

Answer: A. similar chemical properties and characteristics

Explanation:

Pressure = force / area

P = 990 N / (4 × 0.0005 m²)

P = 495,000 Pa

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

Answer:

980 J, B

Explanation:

Given that.

mass of substance, m = 75 g

initial temperature of system, θ1 = 150° C

final temperature of system, θ2 = 250° C

specific heat capacity, c = 0.13 J/gC

Q = mcΔθ, where

Q = quantity of heat required in J

m = mass of substance in G

c = specific heat capacity of substance in J/gC

Δθ = change in temperature °C

Δθ = θ2 - θ1

Δθ = 250° C - 150° C

Δθ = 100° C

now that we have all our values, what we do next is to substitute and apply all in the initial formula given

Q = mcΔθ

Q = 75 * 0.13 * 100

Q = 7500 * 0.13

Q = 975 J

Thus, we can say they amount of heat required to increase the temperature of 75g of gold, from 150° - 250° is 975 J, which is approximately, 980 J.

Option B

Answer:

980 B for plato

Explanation:

Answer:

0.135E-19kgm/s

Explanation:

Using the uncertainty principle, we find

Dp = h / (4π Dx)

= (6.63×10-34Js)/4π(3.90×10-15m) = 0.135×10-19kg m/s

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

             Q_c = m ce ([tex]T_{f}[/tex]-T₀)

Heat released by the largest container is

              Q_a = M ce (T_{i}-T_{f})

how

        Q_c = Q_a

       m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

Answer:

a)143.8 decays/minute

b)0.41 decays/minute

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2=half-life of C-14= 5670 years

t= time taken to decay

Ao= activity of a living sample

A= activity of the sample under study

a)

0.693/5670 = 2.303/1000 log(162.5/A)

1.22×10^-4 = 2.303×10^-3 log(162.5/A)

1.22×10^-4/2.303×10^-3 = log(162.5/A)

0.53 × 10^-1 = log(162.5/A)

5.3 × 10^-2 = log(162.5/A)

162.5/A = Antilog (5.3 × 10^-2 )

A= 162.5/1.13

A= 143.8 decays/minute

b)

0.693/5670 = 2.303/50000 log(162.5/A)

1.22×10^-4 = 4.61×10^-5 log(162.5/A)

1.22×10^-4/4.61×10^-5 = log(162.5/A)

0.26 × 10^1 = log(162.5/A)

2.6= log(162.5/A)

162.5/A = Antilog (2.6 )

A= 162.5/398.1

A= 0.41 decays/minute

Answer:

True

Explanation: A calderas is a large volcanic crater, especially one formed by a major eruption leading to the collapse of the mouth of the volcano.

A Crater is a volcanic crater is an approximately circular depression in the ground caused by volcanic activity. It is typically a bowl-shaped feature within which occurs a vent or vents.

Answer:

The presence of water

Explanation:

Any evidence of water that might appear on the still photo would be a clear indication of life on the planet. This is because scientists believe that for life to thrive elsewhere as it has done here on Earth, it needs water. Water is necessary for fertilization of reproductive cells for some organism, and for others it is where their developing young starts life from. For most, all life biochemical system needs a certain level of moisture to function properly.

Answer:

The total force on the charge at the top of the triangle is approximately

[tex]1.56\,\,10^{-14}\,\,N[/tex]

Explanation:

Please look at the attached image to follow the explanation.

Since all charges are of the same positive value, the force exerted on the top charge by the other two are going to be on the line that joins the top charge with each of the other vertices of the triangle, pointing away from the top charge (illustrated by green vectors of the same length in the image).

We need to find the x and y components of these force vectors in order to find the resultant force by combining the x  components among themselves, and the y components among themselves. Notice that the angle needed is in all cases [tex]60^o[/tex].

The x- components include the cosine of [tex]60^o[/tex], while the y components include the sine of [tex]60^o[/tex].

Notice as well that the x-components cancel each other (they have the same magnitude but point in opposite directions), while the y-components are both of the same value but pointing in the same direction (pointing up).

Then we just need to multiply by two the y component of one of the forces to find this total force.

Now, the magnitude of the forces in question are given by Coulomb's Law:

[tex]F_C=k\,\frac{q_1\,\,q_2}{d^2} =9\,\,10^9\,\frac{2\,\,10^{-12}\,\,2\,\,10^{-12}}{2^2} =9\,\,10^{-15}\,\,N[/tex]

therefore we can calculate what the y-component of this is using:

[tex]F_y=F\,sin(60^o)=9\,\,10^{-15} \,\frac{\sqrt{3} }{2} \\[/tex]

and twice this value becomes:

[tex]2\, \,F_y=(2)\,\,9\,\,10^{-15} \,\frac{\sqrt{3} }{2}=9\,\,\sqrt{3} \,\,10^{-15}\approx 1.56\,\,10^{-14}\,\,N[/tex]

Answer:

The speed of the observer is 2.4 x 10^8 m/s

Explanation:

The standard length of a meter stick is 100 cm

we are to calculate at what speed parallel to the length the length reduces to 60 cm.

This is a relativistic effect question. We know that the length will contract to this 60 cm following the equation below

[tex]l = l_{0}\sqrt{1 - \beta ^{2} }[/tex]

where

[tex]l[/tex] is the new length of 60 cm

[tex]l_{0}[/tex] is the original length which is 100 cm

[tex]\beta[/tex] is the the ratio v/c

where

v is the speed of the observer

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

substituting values, we have

60 = [tex]100\sqrt{1 - \beta ^{2} }[/tex]

0.6 = [tex]\sqrt{1 - \beta ^{2} }[/tex]

we square both side

0.36 = 1 - [tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 1 - 0.36 = 0.64

β = [tex]\sqrt{0.64}[/tex] = 0.8

but β = v/c

v/c = 0.8

substituting value of c, we have

v = 0.8 x 3 x 10^8 = 2.4 x 10^8 m/s

Answer:

1metre

Explanation:

height gained is equal to speed multiplied by angle

Explanation:

we use the formula, Vf=Vi+at

since the cheetah accelerated from rest, it's initial speed is 0, 27=0+a (6.75), a=4 m/s2

Answer: 50 seconds

Explanation:

1km=1000m

Distance/speed=time

1000/20=50

50 seconds

Explanation:

Hey, there!!

Here,

speed (s)= 20m/s

distance (d)= 1km

= 1000m

now,

we have formula,

[tex]t = \frac{d}{s} [/tex]

putting value,

[tex]t = \frac{1000m}{20m/ s } [/tex]

cancelling the like terms,

[tex]t = 50s[/tex]

Therefore, the time is 50s.

Hope it helps...

Answer:

Explanation:

a ) n = 3, m1 = -2

when n = 3 , l = 0 , 1 , 2 .

when l = 2 , one electron may have m = -2 .

two electrons with s = + 1/2 and - 1/2 may have m = -2 .

So only two electrons may have the configuration of n = 3 , m = -2 .

b )  n = 4, l = 3

when , n = 4 and l = 3 , m may have values -3 , -2 , -1 , 0 , 1 , 2 , 3 . Each m have two electrons with s = + 1/2 and - 1/2 .

So altogether there are 7 x 2 = 14 electrons that have configuration of n = 4, l = 3 .

c ) n = 5, l = 3, ml = 2

Similar to case of a ) , only two electrons with s = + 1/2 and - 1/2 may have configuration of n = 5, l = 3, ml = 2

d ) n = 4, l = 1 ml = 0.

In this case also only two electrons may have configuration of

n = 4, l = 1 ml = 0. with s = + 1/2 and - 1/2 .

(a) The maximum number of electrons in an atom that have n = 3, ml = -2 is two electrons.

(b) The maximum number of electrons in an atom that have n = 3, ml = -2 is 14 electrons.

(c) The maximum number of electrons in an atom that have n = 5  l = 3, ml = 2 is two electrons.

(d) The maximum number of electrons in an atom that have n = 4  l = 1 , ml = 0 is two electrons.

The given parameters:

The maximum number of electrons in an atom that have n = 3, ml = -2

The maximum number of electrons in an atom that have n = 4, l = 3

The orbital is 7  which corresponds to f-orbital and the maximum number of electrons = 14;

The maximum number of electrons in an atom that have n = 5  l = 3, ml = 2

The maximum number of electrons in an atom that have n = 4  l = 1 , ml = 0

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Answer:

True

Explanation:

Refraction is the change in the speed, wavelength and direction of a wave as it crosses the boundary between two different media of different densities.

Hence, refraction always involves the movement of a wave from one medium to another. This is the key point to be remembered whether we are discussing refraction in ocean waves or sound waves.