Which of the following best explains Big Bang Theory?

how much energy is needed to heat 2kg of cooking oil with a specific heat capacity of 2000j/kg°c from 20°c to 120°c

mash [69]

Answer:

400kj

Explanation:

h =mass × specific heat capacity × change in temperature

= 2×2000×[120-20]

=2×2000×100

=400000j

=400kj

3 0

3 years ago

Consider a diffraction grating of width 5cm with slits of width 0.0001 cm separated by distance of 0.0002cm 1:what is the corres

sashaice [31]

Answer:

a) $d= 0.0002cm$

b) $m=3$

Explanation:

From the question we are told that:

Width of diffraction grating $W_g= 5cm$

Width of silt $w_s 0.0001 cm$

Distance $d= 0.0002cm$

a)

Generally the grafting element is the distance b/w the silts

$d= 0.0002cm$

Therefore the grafting element is

$d= 0.0002cm$

b)

Generally the equation for diffraction state is mathematically given by

$dsin\theta=m\lambda$

Since

$sin\theta \leq 1$

Therefore

$m \leq \frac{d}{\lambda}$

$m \leq \frac{0.0002}{0.000055}$

$m \leq 3.6$

Therefore orders observed with $\lambda$ is

$m=3$

3 0

3 years ago

8- A 20 KVA, 440/220 single-phase transformer has winding resistance as 0.090 and

Maurinko [17]

The total resistance referred to the HV side will be 0.36 ohm.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The total resistance referred to HV side is;

$\rm \frac{R_H}{R_L} =(\frac{V_P}{V_S})^2 \\\\ R_H=R_L(\frac{V_P}{V_S})^2\\\\ R_H=0.090 \times (\frac{440}{220})^2\\\\ R_H=4 \times 0.090 \\\\ R_H= 0.36 \ ohm$

Hence, the total resistance referred to the HV side will be 0.36 ohm.

To learn more about the resistance refer to the link;

brainly.com/question/20708652

#SPJ1

4 0

2 years ago

A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are pl

Sonbull [250]

Answer:

A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

Explanation:

Parameters given:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between both rods = 4.5cm = 0.045

A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

Where

E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

E₁ = kQ/0.01²

E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²

E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]

E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]

E = 9 * 10⁹ * 15.143* 10⁻⁵

E = 1.36 * 10⁶N/C

B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.02²

E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²

E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]

E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]

E = 9 * 10⁹ * 5.74 * 10⁻⁵

E = 5.17 * 10⁵N/C

C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.03²

E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²

E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]

E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]

E = 9 * 10⁹ * 7.77 * 10⁻⁵

E = 6.993 * 10⁵N/C

4 0

4 years ago

Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f

navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

ΔE = $E_{nf}$ - E₀ = - k²e² / 2m (1 / $n_{f}$ ²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

E = h f = h c / λ

h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

1 / λ = Ry (1/ $n_{f}$ ² - 1 / n₀²)

Let's apply these equations to our case

λ = 821 nm = 821 10⁻⁹ m

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

$E_{nf}$ - E₀ = -13.606 (1 / $n_{f}$ ² - 1 / n₀²) [eV]

Let's look for the energy of some levels

n $E_{n}$ (eV) $E_{nf}$ - E $E_{ni}$ (eV)

1 -13,606 E₂-E₁ = 10.20

2 -3.4015 E₃-E₂ = 1.89

3 -1.512 E₄- E₃ = 0.662

4 -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0

3 years ago