Using the Bohr model, find the ionization energy of the ground He ion.

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

Answer: force is a push or pull that results in the movement of an object ..

Explanation:

Hope this helps you!!!!

Answer;

Cos²စ= I/Io

So

A. Cos²စ = 1/2.2

Cosစ= √1/2.2

စ = cos^-1 0.68

= 47.2°

B.

Cosစ = √1/5.2

စ = ,cos^-1 0.4385

= 64°

C.

Cosစ = √1/12

စ = cos^-1 0.2886

= 73.2°

Answer:

5.448 g/cm³

Explanation:

Density: This is defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³ other sub units are g/cm³, mg/mm³.

From the question,

D = m/πr²h......................... Equation 1

Where D = density, m = mass, r = radius, h = height.

Given: m = 9.58 g, r = 8/2 mm = 4 mm = 0.4 cm, h = 3.5 cm

Substitute this values into equation 1

D = 9.58/(3.14×0.4²×3.5)

D = 5.448 g/cm³

Hence the density of the metal rod is 5.448 g/cm³

Answer:

a

  [tex]n = 23[/tex]

b

  [tex]v = 87377.95 \ m/s[/tex]

Explanation:

From the question we are told that

   The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]

Generally the radius electron orbit  is mathematically represented as

      [tex]r = \frac{61 *10^{-9}}{2}[/tex]

=>   [tex]r = 3.05*10^{-8} \ m[/tex]

This radius can also be represented mathematically  as

      [tex]r = n^2 * a_o[/tex]

Here n is the quantum number and [tex]a_o[/tex] is  the Bohr radius with a value

    [tex]a_o = 0.0529 *10^{-9} \ m[/tex]

So

   [tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]

=>   [tex]n = 23[/tex]

Generally the angular momentum of the electron is mathematically represented as

          [tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]

Here  h is the Planck constant and the value is  [tex]h = 6.626*10^{-34} J \cdot s[/tex]

          m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]

         So

               [tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]

                [tex]v = 87377.95 \ m/s[/tex]

Answer:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

Answer:

After 80 years there will be 3 g of element X remaining

Explanation:

Given;

the half life of element X = 20 year

initial mass of element X = 48 g

a) How much is there after 80 years

0 year --------------------------> 48 g

20 years -----------------------> (48g / 2) = 24 g

40 years ------------------------> 12 g

60 years ------------------------> 6 g

80 years --------------------------> 3 g

Therefore, after 80 years there will be 3 g of element X remaining.

Answer:

B) approaches infinity

Explanation:

The capacitive reactance of an AC capacitor is given by;

[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]

Where;

C is the capacitance

f is the frequency of the ac voltage

[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]

Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.

The correct option is (B) approaches infinity

Answer:

The index of refraction of quartz for violet light is 1.47.

Explanation:

It is given that, a narrow beam of white light is incident on a sheet of quartz.

The beam disperses in the quartz, with red light at an angle, [tex]\theta_r=26.3^{\circ}[/tex] wrt to the normal and violet light traveling at an angle of [tex]\theta_v=25.7^{\circ}[/tex]

The index of refraction of quartz for red light is 1.45.

We need to find the index of refraction of quartz for violet light.

Using Snell's law of red light as follows :

[tex]\mu_a\sin\theta_i=\mu_r\sin\theta_r[/tex]

Here,

[tex]\mu_a[/tex] is the refractive index of air

[tex]\theta_i[/tex] is the angle of incidence

We can find the value of angle of incidence as follows :

[tex]\sin\theta_i=\dfrac{\mu_r \sin\theta_r}{\mu_a}\\\\\sin\theta_i=\dfrac{1.45\times \sin(26.3)}{1}\\\\\theta_i=\sin^{-1}(0.642)\\\\\theta_i=39.79^{\circ}[/tex]

Now again using Snell's law for violet light as follows :

[tex]\mu_a\sin\theta_i=\mu_v\sin\theta_v\\\\\mu_v=\dfrac{\mu_a\sin\theta_i}{\sin\theta_v}\\\\\mu_v=\dfrac{1\times \sin(39.79)}{\sin(25.7)}\\\\\mu_v=1.47[/tex]

So, the index of refraction of quartz for violet light is 1.47.

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation: The Graphite was the material in the passage that had reflected most of the light.

Answer:

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c)  When the frequency ω is doubled, I = 2 A

Explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

[tex]I = \frac{V}{R}[/tex]

(a) When the resistance R is doubled;

[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]

(b)When the peak emf εo is doubled

[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A

Answer:

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Explanation:

Given;

mass of first marble, m₁ = 50g = 0.05 kg

initial velocity of the first marble, u₁ = 2 m/s

mass of second marble, m₂ = 20 g = 0.02 kg

initial velocity of the second marble, u₂ = 0

final velocity of the first marble, v₁ = 1 m/s

Let the final velocity of the second marble, = v₂

Determine the final velocity of the second marble by applying principle of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂

0.1 = 0.05 + 0.02v₂

0.02v₂ = 0.1 - 0.05

0.02v₂ = 0.05

v₂ = 0.05 / 0.02

v₂ = 2.5 m/s

During inelastic collision both objects will move at the same velocity after collision.

During elastic collision both objects will move at different velocities after collision.

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Answer:

Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.

Difference:

Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

Answer:

The time taken is  [tex]t = 0.356 \ s[/tex]

Explanation:

From the question we are told that

  The length of steel the wire is  [tex]l_1 = 31.0 \ m[/tex]

   The  length of the  copper wire is  [tex]l_2 = 17.0 \ m[/tex]

    The  diameter of the wire is  [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]

     The  tension is  [tex]T = 122 \ N[/tex]

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              [tex]t = t_s + t_c[/tex]

Where  [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as

         [tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_s[/tex] is the density of steel with a value  [tex]\rho_s = 8920 \ kg/m^3[/tex]

   So

      [tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_s = 0.235 \ s[/tex]

 And

        [tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as

      [tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]

here  [tex]\rho_c[/tex] is the density of steel with a value  [tex]\rho_s = 7860 \ kg/m^3[/tex]

 So

      [tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]

      [tex]t_c =0.121[/tex]

So  

   [tex]t = t_c + t_s[/tex]

    [tex]t = 0.121 + 0.235[/tex]

    [tex]t = 0.356 \ s[/tex]

Answer: Wavelength of 424 nm  can produce photoelectrons from silver

Explanation:

[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]

[tex]\phi[/tex] = work function = energy of photon

h = Planck's constant =  [tex]6.63\times 10^{-34}Js[/tex]

[tex]\nu_0[/tex] = frequency of the metal

c = speed of light =  [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] =longest  wavelength of the radiation

Now put all the given values in the above formula, we get the wavelength of the photons.

[tex]\lambda=\frac{hc}{\phi}[/tex]

[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex]      ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )

[tex]\lambda=4.24\times 10^{-7}m[/tex]

[tex]1nm=10^{-9}m[/tex]

[tex]\lambda=424nm[/tex]

Here, wavelength of 424 nm  can produce photoelectrons from silver

Explanation:

Given that,

Mass of a box is 2.6 kg

(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.

It means that the position of the object is 3 m i.e. h = 3 m

Work done = Fd

= mgh

So,

[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]

(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,

Change in potential energy = 76.44 J

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

P = 67,000 W

P = 90 hp

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

Answer:

p ≈ f_ objective     Therefore for the object to be in focus it must be close to the focal length

Explanation:

A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length

           p ≈ f_ objective

p distance objetive

Answer:

The magnitude of the acceleration is 5.11 m/s²

Explanation:

Given;

Tension on the rope, T = 164 N

mass of the bucket, m = 11 kg

Weight of the rope is given by;

W = mg = 11 x 9.8 = 107.8 N

According to Newton's second law of motion, the tension on the rope is given by;

T = W - ma

ma = W - T

ma = 107.8N - 164N

ma = -56.2 N

a = -56.2 / m

a = -56.2 / 11

a = -5.11 m/s²

The magnitude of the acceleration is 5.11 m/s²

Answer:

Its not A..

Explanation:

I chose A - was incorrect

Answer:

the sum of all force being applied to an object.

Explanation:

Answer:

The mass of the chunk of gold is 241.25 g

Explanation:

Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:

[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]

Answer:

Physics is the study of matter and the way living things behave everyday....It is related to maths in how we measure the way objects or people do specific things physics has many branches under it that can be helpful too.....Physics teaches us about the world,about the mechanical things we do about air,space,matter about the solar system and about simple machines and things that we do everyday in life.... What we do everyday is related to physics....

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

Answer:

5 n

Explanation:

25 and 25 cancel each other out and 50-45 is 5

Answer:

R_total = 14.57 Ω ,  V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance  

         [tex]R_{eq}[/tex] = ∑ [tex]R_{i}[/tex]

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

         R_B = 20 Ω

Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

          1 / R_BC = 1/20 + 1/15 = 0.116666

          R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

          R_total = R_A + R_BC

          R_total = 6 + 8.57

          R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

           10 = V_a + V_BC

           10 = i R_a + i R_BC = i (R_a + R_BC)

           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

           i = 0.6863 A

The current in the series circuit is constant

          V_BC = i R_BC

          V_BC = 0.6863 8.57

          V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

     Branch C

             V_BC = i R_C5

             i = V_BC / R_C5

             i = 5.8819 / 15

             i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

Answer:

Pressure, P = 3724 Pa

Explanation:

Given that,

Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]

Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]

The density of water, [tex]d_w=1000\ kg/m^3[/tex]

The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]

We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :

[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]

So, the gauge pressure at the bottom of the cylinder is 3724 Pa.

Answer:

The value of the inductance is 4.75 H

Explanation:

Given;

initial current through the inductor, I₁ = 2.4 A

final current through the inductor, I₂ = 0.3 A

duration of change of current, dt = 1.75 s

voltage change of the inductor, V = 5.7 Volts

The voltage change of the inductor is given by;

[tex]V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\[/tex]

Where;

L is the inductance of the coil;

[tex]5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H[/tex]

Therefore, the value of the inductance is 4.75 H