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3 years ago

What is the difference between prokaryotic and eukaryotic?

Physics

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

Prokaryotes are organisms that consist of a single prokaryotic cell. Eukaryotic cells are found in plants, animals, fungi, and protists. They range from 10–100 μm in diameter, and their DNA is contained within a membrane-bound nucleus. Eukaryotes are organisms containing eukaryotic cells.

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The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

astra-53 [7]

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

4 0

2 years ago

A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt

Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m , λ₂ = 2.175 10⁻¹⁰ m, E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

Ef = E₁ + E₂

E₀ = Ef

E₀ = E₁ + E₂

The energy can be found with the Planck equation

E = h f

c = λ f

f = c / λ

E = hc / λ

They indicate that the wavelength of the second photon is

λ₂ = λ₁ +0.130 10⁻⁹

We replace

E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ + λ₁) / λ₁ ( λ₁ + 0.13 10⁻⁹)

1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹ λ₁) = 2 λ₁ + 0.13 10⁻⁹

λ₁² + 0.13 10⁻⁹ λ₁ = 1.25 10⁻¹⁰ λ₁ + 8.125 10⁻²¹

λ₁² + 0.005 10⁻⁹ λ₁ = 8.125 10⁻²¹

λ₁² + 5 10⁻¹² λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

λ₁ = 87.5 10⁻¹² m

λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

λ₁ = 87.5 10⁻¹² m

λ₂ = λ₁ + 0.13 10⁻⁹

λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

E₂ = E₀ –E₁

E₂ = E₀ –hc / λ₁

E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

E₂ = 0.93 10⁻¹⁵ J

E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

E₂ = 5.8 10³ eV

7 0

2 years ago

How do mass and speed affect kinetic energy?

Lunna [17]

The mass affects the kinetic energy because the more the mass the more energy is given to the object and the speed<span> affects by making it go faster and longer, so whenever speed goes up so does energy.</span>

6 0

2 years ago

Read 2 more answers

a circuit has a total resistance of 5 ohms and a current of 0.3 amperes. What voltage must the battery supply?

iren [92.7K]

Answer:

1.5volts

Explanation:

V=IR =0.3×5=1.5volts

5 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$