What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet and the surface tension of honey is 0.052 N/m.
Give your answer to the nearest integer (do NOT use scientific notation).

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.

Answer:

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Explanation:bcz idk

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Explanation:

Pressure = force / area

P = 990 N / (4 × 0.0005 m²)

P = 495,000 Pa

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:

Explanation:

angle of reflection = angle of incidence i = 2θ

angle of refraction r = θ

sin i / sin r = 1.52

sin2θ / sin θ = 1.52

2 sinθ . cosθ / sin θ  = 1.52

2 cosθ = 1.52

cosθ = .76

θ= 41°

Hence angle of incidence = 41°

.

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

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.

Answer:

The amount of energy delivered is 730.84 J

Explanation:

Given;

rms intensity of the electric field, [tex]E_{rms}[/tex] = 6.75 v/m

area of the wall, A = 2.0 cm² = 2.0 x 10⁻⁴ m²

The average intensity of the wave is given  by;

[tex]I_{avg} = c \epsilon_o E_{rms}^2[/tex]

where;

c is the speed of light

ε₀ is permittivity of free space

I = (3 x 10⁸)(8.85 x 10⁻¹²)(6.75)²

I = 0.121 W/m²

Average power delivered, P = I x A

P = 0.121 x 2 x 10⁻⁴

P = 2.42 x 10⁻⁵ W

The amount of energy delivered is calculated as;

E = P x t

E = 2.42 x 10⁻⁵ x 3.02 x 10⁷

E = 730.84 J

Therefore,  the amount of energy delivered is 730.84 J

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:

Explanation:

1 km=1000 m

1 hour =60 min =60*60 sec=3600 sec

Now put 1000 m instead of km and 3600 sec instead of hour in the given expression.

=50 km/hour

=50*1000 m/3600 sec

=500 m/36 sec

=13.89 m/s

Answer:

= 13.89 m/s

Explanation:

     km     1000 m        1 hr

50 ----- x ----------  x ------------  

       hr        1 km      3600 secs

= 13.89 m/s

Answer: v₂ = 5962 km

the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Explanation:

Given that;

Lunar surface is in an altitude h = 43.0 km =  43 × 10³ m

we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²

speed of the space craft when it crashes into the lunar surface , v

decreasing speed of the space craft = 23 m/s

Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r

but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²

HERE r = Rm + h

we substitute

r = 1.74 × 10⁶ m + 43 × 10³ m

= 1.783 × 10⁶ m

On equating these, we have

G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)

v²/r = GM/r²

v = √ ( GM/r)

v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )

v = √ (2750787.9978)

v = 1658.55 m/s

Now since speed is decreasing by 23 m/s

the speed of the space craft into the lunar face is,

v₁ = 1658.55 m/s - 23 m/s

v₁ = 1635.55 m/s

Now applying conversation of energy, we say

1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)

v₂ =  √ [ v₁² + GMe (1/Rm - 1/r)]

v₂ =   √ [ 1635.55²  + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]

v₂ =  √ (2675023.8025 + 67979.24)

v₂ = √(2743003.046)

v₂ = 1656.2 m/s

now convert

v₂ = 1656.2 × 1km/1000m × 3600s/1hrs

v₂ = 5962 km

Therefore the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

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: 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:

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.

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:

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:

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:

The relationship is [tex]E=\frac{ kQ}{d^2}[/tex]

Explanation:

The electric field strength is denoted by the symbol E,

the test charge is denoted be q and the source charge be Q

distance is denoted by d

Then the equation can be rewritten in symbolic form as

Electric field strength is = Force/charge

[tex]E=\frac{F}{q}------1[/tex]

we know that the formula for force is given as

[tex]F= \frac{kQq}{d^2} ------2[/tex]

where [tex]k= 9*10^9 N.m^2/C^2[/tex]

and d is the separation distance between charges

We can insert the expression for Force in equation one

we have

[tex]E= \frac{kQ\sout{q}/d^2}{\xout{q}}--------3[/tex]

We can strike out both qs in the numerator and denominator we have

[tex]E=\frac{ kQ}{d^2}[/tex]

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: A. similar chemical properties and characteristics

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 inductance of one conductor in the unstretched cord is 0.7849 μH

Explanation:

Given;

number of turns of the coil, N = 52 turns

diameter of the coil, d = 1.30 cm

radius of the coil, r = d/2 = 1.30 cm / 2 = 0.65 cm = 0.0065 m

length of the unstretched cord, l = 57.5 cm = 0.575 m

The inductance of one conductor in the unstretched cord is given by;

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

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

A is area of coil, = πr² = π x (0.0065)² = 1.328 x 10⁻⁴ m²

[tex]L = \frac{(52)^2(4\pi *10^{-7})(1.328*10^{-4})}{0.575} \\\\L = 7.849*10^{-7} \ H[/tex]

L = 0.7849 μH

Therefore, the inductance of one conductor in the unstretched cord is 0.7849 μH

Answer:

[tex]f=1.98\times 10^5\ Hz[/tex]

Explanation:

The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by :

[tex]B=1.2\times 10^{-6}\sin [2\pi (\dfrac{z}{240}-\dfrac{10^7t}{8})][/tex] .....(1)

The general equation of the magnetic field wave is given by :

[tex]B=B_o\sin (kz-\omega t)[/tex] ....(2)

Equation (1) is in form of equation (2), if we compare equation (1) and (2) we find that,

[tex]\omega=\dfrac{10^7}{8}[/tex]

We need to find the frequency of the wave. It is given by :

[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{10^7}{8\times 2\pi}\\\\f=1.98\times 10^5\ Hz[/tex]

So, the frequency of the wave is [tex]1.98\times 10^5\ Hz[/tex]

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:

43200c

Explanation:

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:

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