a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron's speed and energy in this state?

Answer:

Explanation:

A and B are 600 m apart . firecrackers explodes and make two waves , sound waves and light waves . they spread  out on all sides . The person sitting in the middle , observes two lights simultaneously and after some time hears two sound simultaneously . It is so because waves coming from A and B covers equal distance to reach the middle point . Same will be the case for sound waves .

But for observer sitting on the other side of A , both light and sound wave will not reach simultaneously . Light wave coming from A will reach earlier because distance covered by light will be less . Similarly sound coming from A will reach earlier . So he / she will observe cracker at A to have exploded earlier than that at B .

Answer:

The thermal conductivity detector (TCD) is a widespread detector dependent on the estimation of the thermal conductivity of a gas. It measures the distinction in heat conductivity between unadulterated carrier gas and carrier gas containing test parts.

Explanation:

When you have two parallel tubes both containing gas and coils being heated, thermal conductivity occurs. The gases are inspected by comparing the rate of heat loss in the heated coils in the gas. One of the cylinders has a reference gas and the other cylinder contains the sample to be tested. A thermal conductivity detector will sense the change in the thermal conductivity and compares it to a reference stream of carrier gas. An analyte washed from the section causes the reduction of the thermal conductivity of the effluent and produces a detectable signal. The thermal conductivity of most compounds is less than the thermal conductivity of hydrogen and helium.

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:

A. 1/50π V

Explanation:

Given;

magnetic flux through the coil, φm = (1/50π) cos 100πt

t = 1/200 s

The magnitude of the induced EMF is given by;

[tex]EMF = \frac{d \phi_m}{dt} \\\\EMF =\frac{d}{dt} (1/50 \pi)cos \ 100\pi t)\\\\EMF = (1/50 \pi)cos \ 100\pi \\\\EMF =(1/50 \pi)cos \ 100 *180\\\\ EMF =(1/50 \pi)cos \ 18000\\\\ EMF =(1/50 \pi) (1)\\\\EMF =(1/50 \pi) \ V[/tex]

Therefore, the magnitude of the induced EMF is 1/50π V

The correct option is "A. 1/50π V"

Answer:

The  velocity is  [tex]v_t = 0.02175 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass of the bullet is  [tex]m_b = 0.024 \ kg[/tex]

    The initial speed of the bullet is  [tex]u_b = 1200 \ m/s[/tex]

   The mass of the target is  [tex]m_t = 320 \ kg[/tex]

    The  initial velocity of target is  [tex]u_t = 0 \ m/s[/tex]

    The  final velocity of the bullet is  is  [tex]v_b = 910 \ m/s[/tex]

Generally according to the law of momentum conservation we have that

      [tex]m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t[/tex]

=>   [tex]0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t[/tex]

=>    [tex]v_t = 0.02175 \ m/s[/tex]

Answer:

If you position yourself between the center of curvature and the focal point of the mirror, you will see an enlarged image of your face.

Explanation:

A concave mirror is a curved mirror. The position of an object placed at any point in front of the mirror and the corresponding image formed by the concave mirror are depicted in appropriate ray diagrams.

If I place my face between the centre of curvature and the principal focus of the concave mirror, the image formed will be enlarged, real and inverted.

Answer:

The  value  is   [tex]m \approx 310[/tex]

Explanation:

From the question we are told that

     The  focal length of the objective is  [tex]f_o = 1.0 \ cm[/tex]

    The  focal length of the eyepiece is  [tex]f_e = 2.0 \ cm[/tex]

    The  tube length is  [tex]L = 25 \ cm[/tex]

Generally the magnitude of the overall magnification is mathematically represented as

            [tex]m = m_o * m_e[/tex]

Where  [tex]m_o[/tex] is the objective magnification which is mathematically represented as

        [tex]m_o = \frac{L}{f_o }[/tex]

=>      [tex]m_o = \frac{25}{1 }[/tex]

=>      [tex]m_o = 25[/tex]

[tex]m_e[/tex] is the eyepiece magnification which is mathematically evaluated as

     [tex]m_e = \frac{L }{f_e }[/tex]

     [tex]m_e = \frac{25 }{ 2}[/tex]

      [tex]m_e = 12.5 \ cm[/tex]

So

    [tex]m = 25 * 12.5[/tex]

     [tex]m \approx 310[/tex]

Answer:

  ρ_liquid = 1,778 10³ kg/m³

Explanation:

In a wire the speed of the wave produced is

          v = √T/μ

wave speed is also related to frequency and wavelength

           v = λ f

as they indicate that the frequency is the fundamental, there must be a single antinode in the center

         L = 2λ

         λ = 2L

Using the translational equilibrium equation

         T -W = 0

         T = W

let's substitute

         2L f = √ W /μ

         μ = W / 4 L² f²

         μ = 164 / (4 3² 42²)

         μ = 2.5825 10⁻³ kg / m

this is the linear density of the wire

Now let's analyze when the rock is submerged in a liquid, the thrust acts on the rock

             B = ρ g V

when writing the equilibrium equation we have

              T + B -W = 0

              T = W - B

               T = W - ρ_liquid g V                    (1)

to find the volume of the rock we use the concept of density

              ρ_body = M / V

              V = M /ρ_body

              W = M g

              V = W / g ρ_body

              V = 164 / (9.8 3200)

              V = 0.00523 m³

               V = 5.23 10⁻³ m³

The speed of a wave on a string is

              v = √ T /μ

speed is also related to wavelength and frequency

              v = λ f

indicate that the frequency formed is the fundamental one, therefore it has a single antinode in the center and two nodes at the fixed points

             L = λ/ 2

             λ= 2L

we substitute and look for tension

             2L f = √T /μ

              T = 4L² f² μ

              T = 4 3² 28² 2.5825 10⁻³

              T = 72.888 N

We already have the volume of the rock and the tension on the rope, we can substitute in equation 1 and find the density of the liquid

               T= W – ρ_liquid g V  

               ρ_liquid = (W -T) / gV

               ρ_liquid = ( 164 – 72,888) / ( 9,8 5,23 10⁻³)

               ρ_liquid = 1,778 10³ kg/m³

Explanation:

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Answer:

no its a negative

Explanation:

because they both are positive

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:

(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

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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 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:

a) 1 & 2The pressure at the surface of water and just outside the hole are both atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

3. P(bottom) = 2987.04 Pa

b) v = 1.0844 m/s

c) v₁  = 0.11 m/s

Explanation:

a)

1) The pressure at the surface of water is same as the atmospheric pressure { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

2) The pressure at just outside the hole is also same as the atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}

3) At the bottom of the container, the gauge pressure is given by:

P(bottom) = pgh

where h = 12in = 0.3084m

P(bottom) = 1000 kg/m³ × 9.8 m/s² × 0.3084m

P(bottom) = 2987.04 Pa

b)

The velocity at which the water leaves the hole can be obtained from Bernouilli's equation. Point 1 is at the surface of water, point 2 is just outside the hole;

p₁ + 1/2pv₁² + pgh₁ = p² + 1/2pv₂² + pgh₂

Pressures on both sides are the same and the velocity of the water at the surface is approximately zero (since the hole's cross sectional area is much smaller than the container's). The difference in depths h1 - h2 is just the height of the water level

so

pgh =  1/2pv₂²

v = √(2gh)

we substitute

v = √( 2 × 9.8 m/s² × 0.06 m

v = 1.0844 m/s

c)

Now the velocity of the water level can't be neglected, we can use the continuity equation:

v₂ = (A₁/A²)V₁

so Back to Bernouilli's:

1/2pv₁² + pgh₁ = 1/2 (10v₁)² + pgh₂

phg = 99/2pv₁²

v₁ = √(2/99gh)

v₁  = √(2/99 × 9.8 m/s² × 0.06 m)

v₁  = 0.11 m/s

Answer:

A is walking

B is stationary

C is running

D is walking

E is stationary

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

[tex]P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}[/tex]

Where;

F₂ is the force on big piston

[tex]F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N[/tex]

Therefore, the force applied on the big piston is 1306.67 N

Answer:

ocean-continental convergence

A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?

Choose: 100 kilometers/hour south

Answer:

100 kilometers/hour its d

Answer:

Explanation:

distance travelled s = 10 cm

speed v = 1.4 x 10⁶ m /s

v² = u² + 2as

u = 0

v² = 2as

( 1.4 x 10⁶ )² = 2 x a x .10

a = 9.8 x 10¹² m /s²

force on proton = mass x acceleration

= 1.67 x 10⁻²⁷ x 9.8 x 10¹²

= 16.366 x 10⁻¹⁵ N .

If magnitude of electric field be E

force on proton

= E x charge on proton

= E x 1.6 x 10⁻¹⁹

E x 1.6 x 10⁻¹⁹ = 16.366 x 10⁻¹⁵

E = 10.22 x 10⁴ N/C

The direction of electric field will be positive x - direction .

b )

Speed of proton = 1.4 x 10⁶  m /s .

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: white light

Explanation:

if you mix red and blue light together, you get magenta and when you mix blue and green light together, you get cyan and when you mix red and green light together, you get yellow but when you mix all three primary colors of light together, you end up with white light or as we see it.

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:

The wavelength is  [tex]\lambda_R = 649 *10^{-9}\ m[/tex]

Explanation:

From the question we are told that

   The wavelength of the red laser is  [tex]\lambda_r = 632.8 \ nm = 632.8 *10^{-9}\ m[/tex]

    The spacing between  the fringe is  [tex]y_r = 6.00\ mm = 6.00*10^{-3} \ m[/tex]

   The spacing between  the fringe for smaller laser point  is  [tex]y_R = 6.19 \ mm = 6.19 *10^{-3} \ m[/tex]

      Generally the spacing between  the fringe is mathematically represented as

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

Here  [tex]D[/tex] is the distance to the screen

    and  d is the distance of the slit separation

Now for both laser red light light and  small laser  point  D and  d are same for this experiment

So

         [tex]\frac{y_r}{\lambda_r} = \frac{D}{d}[/tex]

=>      [tex]\frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}[/tex]

Where [tex]\lambda_R[/tex]  is the wavelength produced by the small laser pointer

  So

           [tex]\frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}[/tex]

=>       [tex]\lambda_R = 649 *10^{-9}\ m[/tex]

Answer:mile

Explanation: heres a hint think aboyt the distance between your house to school

Answer:

BS33

Explanation:

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

Answer:

The angle is  [tex]\theta = 6.93 *10^{1}[/tex]

Explanation:

From the question we are told that

   The intensity of  light emerging from the second polarizer is  [tex]I_2 = \frac{1}{8} * I_1[/tex]

Now  the light emerging from the first polarizer is  [tex]I_1 = \frac{I_o}{2}[/tex]

 Where [tex]I_o[/tex] is the intensity of the unpolarized light

 Now according to Malus law the intensity of light emerging from the second polarizer is mathematically represented as

       [tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]

=>   [tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]

=>   [tex]\theta = cos^{-1}[0.3536][/tex]

=>   [tex]\theta = 69.3^o[/tex]

=>  [tex]\theta = 6.93 *10^{1}[/tex]

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]

Answer:

The magnetic field circles the wire in a counter-clockwise direction.

Explanation:

This is by using Fleming right hand rule which says If you point your pointer finger in the direction the positive charge is moving, and then your middle finger in the direction of the magnetic field, your thumb points in the direction of the magnetic force pushing on the moving charge.

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: