A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

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

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

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

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

Someone please answer this, ill give you brainliest and your earning 50 points.

Tema [17]

All of the following are non-renewable resources except

O natural gas

O oil

O minerals

O water ✓ 

Water is a renewable source because evaporation and condensation takes place everytime on our planet

3 0

2 years ago

Read 2 more answers

While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the

BartSMP [9]

In the experiment of free fall bob released a bag of mass 1 lb

so here we can say that initial speed of the bag is Zero

time taken by the bag to free fall is given as

t = 1.5 s

also the acceleration of free fall is given as

a = 9.8 m/s^2

now we will use kinematics equation here for finding the distance of free fall

$d = v_i * t + \frac{1}{2} at^2$

$d = 0 + \frac{1}{2}*9.8* 1.5^2$

$d = 4.9 * 2.25$

$d = 11.025 m$

so the bag will fall down by total distance of 11.025 m from its initial released position.

3 0

3 years ago

What happens if a mid-ocean ridge occurs on land

Gnesinka [82]

Answer:

Despite being such prominent feature on our planet, much of the mid-ocean ridge system remains a mystery. While we have mapped about half of the global mid-ocean ridge in high resolution, less than one percent of the mid-ocean ridge has been explored in detail using submersibles or remotely operated vehicles. so therefore we do not have enough information about them to know what will happen

Explanation:

A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary. Mid-ocean ridges occur along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing enormous volcanic eruptions of basalt. The speed of spreading affects the shape of a ridge slower spreading rates result in steep, irregular topography while faster spreading rates produce much wider profiles and more gentle slopes.

4 0

3 years ago

A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L

GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒ L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁

⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂

⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

3 0

3 years ago