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3 years ago

What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?

1 answer:

4 0

Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant

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2. Sara y Antonio son mellizos. Cuando nacieron, Sara pesaba 600 gramos más que Antonio. Sus pesos ya se han igualado, gracias a

valkas [14]

Answer:

8,25 creo

Explanation:

nose no estoy segura

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3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

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3 years ago

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Tema [17]

<span>Kinetic energy increases and potential energy decreases.
</span>

6 0

3 years ago

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In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

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3 years ago

18)

Nina [5.8K]

Number 18 is C, i thin

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3 years ago

Read 2 more answers