Which image shows a nonrenewable resource? sun fire can plant

Monica [59]

In real machines, efficiency is always less than 1 .

6 0

3 years ago

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

Sergeeva-Olga [200]

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x 10⁻⁹ m

λ = 319.83 nm

Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

5 0

3 years ago

What is providing the centripetal force in the following examples?

umka2103 [35]

A yo-yo swung in a circle.

8 0

3 years ago

Read 2 more answers

Which of the following has the largest momentum?

san4es73 [151]

Momentum = mass x velocity

Thus Option A is the correct answer

Momentum (dog) = 10 kg x (0.447 x 30) m/s

= 134.1 Kg m/s

Momentum ( bullet) = 0.02 kg x (0.447 x 800) m/s

= 7.152 Kg m/s

Momentum ( truck) = 0, as v = 0

tightrope has both low mass and low speed, thus its momentum will be low

4 0

3 years ago

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.96 x 10³ eV

B)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 4.19 x 10⁴ eV

C)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 3.73 x 10⁹ eV

8 0

2 years ago