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kiruha [24]
3 years ago
15

Question 3

Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

oinkhbbkjfcbjtfcvbnklo0756

You might be interested in
3. Two Metra trains approach each other on separate but parallel tracks. Train A has a speed of 90 km/ hr, train B has a speed o
Elden [556K]

Answer:he trains take 57.4 s to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

The relative velocity of the train A with respect to B is given by,

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

Substitute 2.71 km for d and 170 km/h for

Express the time in seconds.

Thus, the trains cross each other in 57.4 s.

Explanation:

8 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed
Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

sin\theta = \frac{\lambda}{a}

              == \frac{633 *10^{-9}}{.360*10^{-3}}

           \theta = 0.100

y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m

with of central fringe  = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = w' = y_2 -y_1

calculation for y_2

sin\theta = 2\frac{\lambda}{a}

              = = 2*\frac{633 *10^{-9}}{.360*10^{-3}}

             \theta  = 2*0.100 = 0.200

y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m

w' = 0.0130  -6.54 *10^{-3}

w' = =6.46*10^{-3] m

6 0
3 years ago
2 strings both vibrate at exactly 220 Hz. The tension in one of them is then decreased sightly. As a result, 3 beats per second
MAVERICK [17]

Explanation:

Given that,

2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.

The tension in one of them is then decreased sightly, then f_2 will decrese.

Beat frequency, f=f_1-f_2

3=220-f_2

f_2=217\ Hz

So, the new frequency of the string is 217 Hz. Hence, this is the required solution.

3 0
3 years ago
Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul
timofeeve [1]

Answer:

a)   K_{max} = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

      K_{max}= h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

     c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

     Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

     E = 2,486 10⁻¹⁸ J

Let's reduce to eV

     E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

     E = 15.5 eV

Now we can calculate the kinetic energy

     K_{max}= h c / f - fi

      K_{max} = 15.5 -13.6

     K_{max} = 1.9 eV

b)     Extra energy = 10.2 eV

The total kinetic energy of electrons is

       Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

      ΔE = E photon - Total kinetic energy electron

      ΔE = 15.5 - 12.1

      ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

     E_{n} = 13.606 / n2

     n = √ 13.606 / En

     n = √ (13606 / 3.4)

     n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

8 0
3 years ago
A 79 kg person sits on a 3.7 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.3 cm in diameter
vampirchik [111]
I'll assume that the chair has four legs.

Since the chair weights 3.7 kg by itself, it will weigh (79+3.7)=82.7 kg with the person sitting on it. And each of the chair's legs will take about (82.7/4)=20.675 kg.

Each leg touches the floor in a circle with 1.3cm diameter. The area of that circle is about (3.14*(1.3/2)^2)=1.327 cm^2.

Pressure is measured by force per area. So, the pressure from each leg is about 20.675kg / 1.327cm^2. That simplifies to 15.58 kg/cm^2.
7 0
3 years ago
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