Answer:

**Answer:**

**Ionization potential of C⁺⁵ is 489.6 eV.**

**Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m. **

**Explanation:**

The ionization potential of hydrogen like atoms is given by the relation :

.....(1)

Here *E* is ionization potential, *Z* is atomic number and *n* is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for *Z* and 1 for *n* in the equation (1).

* E = ***489.6 eV**

The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

= 5.45 x 10⁷

**λ = 1.83 x 10⁻⁸ m**

Which of the following statement(s) istrue about clipper circuitsSelect one:O a. It converts a.c to d.c alwaysO b. The output is clipped of someportion of input signalO c. It converts d.c to a.cO d. None of these

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.

**Answer:**

**Explanation:**

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final = 1/2 I₂ ω₂²

= 1/2 (I₁ / 2) (2ω)²

= I₁ ω²

c )

Change in rotational kinetic energy

= I₁ ω² - 1/2 I₁ ω²

= + 1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

**Answer:**

**Explanation:**

Let the two capacitance are

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So --------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So

-----EQN

On solving eqn 1 and eqn 2

**Answer:**

v = 4.1 m / s

**Explanation:**

Velocity is defined by the relation

v =

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v =

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The **Speed of a Particle** is 4.1 meters per second.

The position of a particle can be represented by a **linear equation** of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the **derivative of the position** equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the **velocity of the particle**.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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The **anchoring force** needed to hold the **elbow **in place is 839.5 N.

To determine the anchoring force **needed to hold** the elbow in **place**, use the following steps:

Apply the **conservation of momentum** equation to the elbow:

∑F = ˙m(V₂ - V₁)

where:

˙m = mass flow rate

V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively

F = anchoring force

The **mass flow rate** is given by:

˙m = ρAV

where:

ρ = density of water (1000 kg/m³)

A = cross-sectional area of the elbow

V = velocity

The **velocities** at the inlet and outlet of the elbow can be calculated using the following equations:

V₁ = A₁V₁

V₂ = A₂V₂

where:

A₁ and A₂ = **cross-sectional areas** at the inlet and outlet of the elbow, respectively

Calculate the momentum flux correction factor at the inlet and outlet of the elbow:

β₁ = 1.03

β₂ = 1.03

Substitute all of the above equations into the conservation of momentum equation:

F = ˙m(V₂ - V₁)

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

Calculate the velocity at the **inlet **of the elbow:

V₁ = A₂V₂/A₁

V₁ = (25 cm²/150 cm²)(V₂)

V₁ = 1/6 V₂

Substitute the above equation into the conservation of momentum equation:

F = ρA₁V₁²β₁ - ρA₂V₂²β₂

F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)

F = ρV₂²(1.03)(1/36 A₁ - A₂)

Calculate the **anchoring force**:

F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)

F = 839.5 N

Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.

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The **anchoring force** needed to hold the elbow in place is 932 N.

The anchoring force needed to hold the elbow in place is the sum of the following forces:

The force due to the change in **momentum **of the water as it flows through the elbow.

The force due to the weight of the elbow and the water in it.

The force due to the buoyancy of the water in the elbow.

The force due to the **change **in momentum of the water can be calculated using the momentum equation:

F = mΔv

where:

F is the force

m is the mass of the fluid

Δv is the change in velocity of the fluid

In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:

m = ρAv

where:

ρ is the **density **of the fluid

A is the cross-sectional area of the pipe

v is the velocity of the fluid

The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:

P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2

where:

P1 is the pressure at the inlet

v1 is the **velocity **at the inlet

P2 is the pressure at the outlet

v2 is the velocity at the outlet

In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:

A1v1 = A2v2

where:

A1 is the cross-sectional area at the inlet

A2 is the cross-**sectional **area at the outlet

The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:

W = mg

where:

W is the weight

m is the mass

g is the acceleration due to gravity

The force due to the **buoyancy **of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:

B = ρVg

where:

B is the buoyancy

ρ is the density of the fluid

V is the volume of the fluid displaced

g is the acceleration due to gravity

The volume of the fluid **displaced **by the elbow is equal to the volume of the elbow.

Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.

F_anchor = F_momentum + F_weight - F_buoyancy

where:

F_anchor is the anchoring force

F_momentum is the force due to the change in momentum of the water

F_weight is the force due to the **weight **of the elbow and the water in it

F_buoyancy is the force due to the buoyancy of the water in the elbow

Plugging in the values for each force, we get:

F_anchor = 1030 N - 490 N + 392 N = 932 N

Therefore, the anchoring force needed to hold the elbow in place is 932 N.

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**Answer:**

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

**Explanation:**

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

Here h is the Planck's constant, c is the speed of light, is the wavelength of the light and W the work function of the element:

Now, we calculate the wavelength for the new maximum kinetic energy:

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

The **magnitude **of the **speed **is 83.0325 m\s, the **direction **is 62.7 degrees, and the **fraction **of **kinetic energy **lost is 0.895.

The **collision **is the phenomenon when **two **objects come in **direct **contact with **each **other. Then both the **bodies **exert **forces **on each other.

The **mass**, **angle**, and **velocity **of the **first **object are 5.12 g, 21.3°, and 239 m/s.

And the **mass**, **angle**, and **velocity **of the **second **object be 3.05 g, 15.4°, and 282 m/s.

The **momentum **(P₁) before a **collision **will be

The **momentum **(P₂) after a **collision **will be

Applying **momentum conservation**, we have

...1

...2

From **equations **1 and 2, we have

From **equation **1, we have

Then the **change **in **kinetic energy**, we have

The **fraction **of **kinetic energy **lost will be

More about the **collision **link is given below.

**Answer:**

Detailed solution is given below