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4 years ago

Which is a characteristic of all waves?

Physics

2 answers:

juin [17]4 years ago

7 0

Answer:à

Explanation:waves carry energy in the direction in which they move

kramer4 years ago

4 0

Characteristics of all waves:

<em>All waves carry energy. </em>

All waves move through space.

All waves move through matter, from particle to particle.

<em>All waves transfer energy parallel to the direction of wave mo tion.</em>

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The surface tension of water was determuned in a laboratory by using the drop weight method. 100 drops were released from a bure

lana [24]

Answer:

The surface tension of the water is 6.278×10⁻² N/m

error = 13.65%

Explanation:

The surface tension of water is given by

$\gamma = \frac{F}{L}$

Where F is the force acting on water and L is the length over which is force is acted.

We are given the mass of 100 droplets of water

M = 3.78 g

n = 100

The mass of 1 droplet is given by

$m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg$

The force acting on a single droplet of water is given by

$F = m \cdot g$

Where m is the mass of water droplet and g is the acceleration due to gravity

$F = 3.780\times10^{-5} \cdot 9.81$

$F = 3.708\times10^{-4} \: N$

The circumferential length of the droplet is given by

$L = \pi \cdot d$

Where d is the diameter

$L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m$

Now we can find out the required surface tension of the water

$\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m\\\\$

Therefore, the surface tension of the water is 6.278×10⁻² N/m

The tabulated value of the surface tension of water at 20 °C is given by

$\gamma_t = 0.0727 \: N/m$

The percentage error between tabulated and calculated surface tension is given by

$error = \frac{\gamma_t - \gamma }{\gamma_t}$

$error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\%$

$$ error = 13.65 \%$

7 0

3 years ago

The orbital velocity of the Earth about the sun is 30 km/s. If the Earth were suddenly stopped in its tracks, it would simply fa

soldi70 [24.7K]

Answer:

Planet will crash on to the Sun if the tangential velocity becomes zero and Rocket should be fired from Earth's orbit is at 30 m/s and in opposite direction to the Earth orbits the Sun

Explanation:

The orbital velocity of the Earth about the sun is 30 km/s. If we shoot a rocket with 30 km/s with respect to Earth in the opposite direction. Then the two velocity vectors cancel. The resultant velocity would be zero with respect to the Sun. resulting velocity is called as tangential velocity.

Planet will crash on to the Sun if the tangential velocity becomes zero and Rocket should be fired from Earth's orbit is at 30 m/s and in opposite direction to the Earth orbits the Sun

4 0

4 years ago

A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?

Aleksandr-060686 [28]

Answer:

1.2 102j

Explanation:

because it is the most important for me

3 0

3 years ago

Based on the data, which of the following explains the relationship between the distance of the planet from the Sun and its plan

Fofino [41]

D the planets with the greatest

7 0

3 years ago

Read 2 more answers

You have a great summer job working in a cancer research laboratory. Your team is trying to construct a gas laser that will give

Alchen [17]

Answer:

ΔE = 7.559 eV , λ = 1,645 10⁻⁷ m

Explanation:

For this exercise we can use the Bohr model for ionized atom with only one free electron,

r_n = n² a₀ / Z

E_n = -13,606 Z² / n²

Where a₀ is the Bohr radius of the hydrogen atom (a₀ = 0.0529 nm), Z is the atomic number of the atom under study and 13.606 eV is the energy of the ground state of Hydrogen.

In our case the Helium atom has two protons Z = 2

let's calculate the quantum number and the energy of each orbit

r_n = 0.30 nm

n₁ = √ (r_n Z / a₀)

n₁ = √ (0.30 2 / 0.0529)

Note that we do not have to reduce the radius since they are all in nanometers

n₁ = 3.3

since n is an integer we approximate it to

n₁ = 3

r_n = 0.20 nm

n₂ = √ (0.2 2 / 0.0529)

n₂ = 2.7

To approximate this value we must assume that there could be some error in the medicinal radio,

n₂ = 2

having the quantum numbers of the two radius we can calculate their energy

E₃ = - 13,606 2²/3²

E₃ = - 6.047 eV

E₂ = -13.606 2²/2²

E₂ = -13.606 eV

the energy of the emitted photon is

ΔE = E₃ - E₂

ΔE = -6.047 + 13.606

ΔE = 7.559 eV

You do not indicate in the exercise if you want the energy or the wavelength of the photon,

to find the wavelength We use the Planck relation

E = h f

c = λ f

E = h c /λ

λ = h c / E

we must reduce the energy to the SI system

E = 7.559 ev (1.6 10⁻¹⁹ J / 1eV) = 12.09 10⁻¹⁹ J

λ = 6.63 10⁻³⁴ 3 10⁸ / 12.09 10⁻¹⁹

λ = 1,645 10⁻⁷ m

6 0

4 years ago