Maria was returning from her work when, when entering a street, she came across two unknown boys, who began to give praise to the girl:
_ You are the most beautiful girl seen in this world, said the first young man.
_ The moon is jealous of you! Said the second young man.
The girl went on her way and left.
What's wrong with this situation?

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:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

[tex]A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m[/tex]

(c) the maximum speed attained by the object during its motion

[tex]E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s[/tex]

a. The total energy of the toy at any point in its motion is 0.0416 Joules.

b. The amplitude of the motion is equal to 0.0167 meter.

c. The maximum speed attained by the toy during its motion is 0.577 m/s.

Given the following data:

a. To find the total energy of the toy at any point in its motion:

Mathematically, the total energy of an object undergoing simple harmonic motion (SHM) is given by:

[tex]E = \frac{1}{2} MV^2 + \frac{1}{2} kx^2[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]E = \frac{1}{2} \times 0.25 \times 0.400^2 + \frac{1}{2} \times 300 \times 0.0120^2\\\\E = 0.125 \times 0.16 + 150 \times 0.000144\\\\E=0.02+0.0216[/tex]

E = 0.0416 Joules.

b. To find the amplitude of the motion, we would use this formula:

[tex]A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2 \times 0.0416}{300} } \\\\A = \sqrt{2.77 \times 10^{-4}}[/tex]

A = 0.0167 meter.

c. To find the maximum speed attained by the object during its motion:

[tex]V_{max} = \sqrt{\frac{2E}{M} } \\\\V_{max} = \sqrt{\frac{2 \times 0.0416}{0.25} } \\\\V_{max} = \sqrt{2.77 \times 10^{-4}}[/tex]

Maximum speed = 0.577 m/s

Read more: https://brainly.com/question/14621920

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.

Answer:

D. The sphere the disk and the hoop

Explanation:

This is because the sphere has inertial of

2/5mR²

Disk 1/2mR²

Hope mR²

So these are moment of inertial which is resistance or opposition to rotation so since the sphere has a smaller moment to inertial it will move faster and reach the ground first then the disk then the hoop in that order

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

Learn more about quantum number of electrons here: https://brainly.com/question/11575590

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:

ocean-continental convergence

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:

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:

no its a negative

Explanation:

because they both are positive

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:

a. 0 T b. 1.03 × 10⁻⁴ T c. 2.29 × 10⁻⁴ T

Explanation:

a. Using ampere's law ∫B.ds = μ₀i

for ra = r/2, i = 0 (since no current is enclosed) and

∫B.ds = B∫ds = B(2πr/2) = Bπr

So, Bπr = 0

B = 0 T

b. rb = 5r/4

WE find the current enclosed between r = r and r = 5r/4. The total current density in the holow thick-walled conducting cylinder is J = i/[π(3r/2)² - πr²] = i/[9πr²/4 - πr²] = i/8πr²/4 = i/2πr².

Since the current density is constant, we find the current, i' enclosed between  r = r and r = 5r/4. J = i'//[π(5r/4)² - πr²] = i'/[25πr²/16 - πr²] = i'/9πr²/16 = 16i'/9πr²

So, i/2πr² = 16i'/9πr²

i' = 9i/32

Using ampere's law ∫B.ds = μ₀i'

∫B.ds = = B∫ds = B × 2π(5r/4) = 5Bπr/2

5Bπr/2 = μ₀(9i/32)

B = 9μ₀i/32πr × 2/5

B = 9μ₀i/80πr

Substituting r = 4.90 mm = 4.90 × 10⁻³ m and i = 11.2 A, we have

B = 9μ₀i/80πr

= 9 × 4π × 10⁻⁷ H/m × 11.2 A/80π(4.90 × 10⁻³ m)

= 403.2/392 × 10⁻⁴ T

= 1.029 × 10⁻⁴ T

≅ 1.03 × 10⁻⁴ T

c. rc = 2r

Using ampere's law ∫B.ds = μ₀i

∫B.ds = = B∫ds = B × 2π(2r) = 4Bπr

4Bπr = μ₀i  (since i = current enclosed = 11.2 A)

B =  μ₀i/4πr

= 4π × 10⁻⁷ H/m × 11.2 A/4π(4.90 × 10⁻³ m)

= 2.286 × 10⁻⁴ T

≅ 2.29 × 10⁻⁴ T

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:

edqr3fr =wdmmwfmnl[=kvlfkgvb

Explanation:bcz idk

Me BeLIves iT iS ZeRo So A

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:

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:

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:

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

Explanation:

Pressure = force / area

P = 990 N / (4 × 0.0005 m²)

P = 495,000 Pa

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:

0.083 V

Explanation:

The equation of the reaction is;

Pb(s) + 2H^+(aq) -------> Pb^2+(aq) + H2(g)

E°cell = E°cathode - E°anode

E°cathode = 0.00 V

E°anode = -0.13 V

E°cell = 0.00-(-0.13)

E°cell = 0.13 V

Q= [Pb^2+] pH2/[H^+]^2

Q= 0.1 × 1/[0.05]^2

Q= 0.1/0.0025

Q= 40

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

Ecell= 0.13 - 0.0592/2 log (40)

Ecell = 0.13 - 0.047

Ecell= 0.083 V

Answer:

the speed acquired by the man is: [tex]0.00308\,\,\frac{m}{s}[/tex]

Explanation:

Use conservation of linear momentum to solve this problem, and make sure you start by converting the 63 g stone mass into kilograms = 0.063 kg.

Now from conservation of linear momentum, the momentum imparted to the stone (which is the product of the stone's mass time its speed) must equal in magnitude that of the 90 kg man. Then we need to find the unknown speed of the man:

[tex]m_1\.v_1=m_2\,v_2\\0.063\,(4.4)\,\frac{kg\,m}{s} =v_2\,(90\,\,kg)\\0.2772\,\frac{kg\,m}{s}=v_2\,(90\,\,kg)\\v_2=\frac{0.2772}{90} \,\frac{m}{s} \\v_2=0.00308\,\,\frac{m}{s}[/tex]

therefore the speed acquired by the man is: [tex]0.00308\,\,\frac{m}{s}[/tex]

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:

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:

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:

the orbit resulting an ELIPSE

Explanation:

Harmonic motion is described by the expression

          x = A cos (wt +Ф₁)

In this exercise it is established that in the y axis there is also a harmonic movement with the same frequency, so its equation is

        y = B cos (wt + Ф₂)

the combined motion of the two bodies can be found using the Pythagorean theorem

        R² = x² + y²

        R² = [A² cos² (wt + Ф₁) + B² cos² (wt + Ф₂)]

to simplify we can assume that the phase in the two movements are equal

        R = √(A² + B²) cos (wt + Ф)

        If the two amplitudes are equal we have a circular motion, if the two amplitudes are different in elliptical motion, the amplitudes of the two motions are

circular    R² = (A² + A²)

elliptical   R² = (A² + B²)

We see from the last expression that the broadly the two axes is different, so the amplitude is an ellipse.

By which the orbit resulting an ELIPSE

Answer:

1metre

Explanation:

height gained is equal to speed multiplied by angle