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bulgar [2K]
3 years ago
8

What is a translucent object?​

Physics
2 answers:
jarptica [38.1K]3 years ago
4 0
A translucent object allows light to travel through its material.
labwork [276]3 years ago
3 0

an object that isn't completely transparent or opaque kinda the middleground

it allows light through but it's not clear

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How might observations of the sun's outermost layers reveal what's happening in the interior? and how could this information be
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A raging activity can be found in t<span>he Sun's interior, with pressure waves being produced and travelling back and forth, from the core to the surface and back to the core. By looking closely at the 'surface' we can see these "ripples".  It gives us an idea of how dense the material was that the waves passed through. In a way, this can help to predict solar storms in the future.</span>
8 0
4 years ago
Read 2 more answers
Un Esquiador Inicia un Salto Horizontal con una velocidad Inicial de 25 m/s. La Altura al Final de la Rampa es de 80 m del punto
Evgesh-ka [11]

Responder:

Explicación:

Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;

T = 2Usin theta / 2

theta = 90 grados

U = 25 m / s

T = 25sin90 / 2 (9,8)

T = 25 / 19,62

T = 1,27 segundos

Por lo tanto, los cielos usarán 1.27 segundos en el aire.

La distancia horizontal es el rango;

Rango R = U√2H / g

R = 25√2 (80) /9,8

R = 25√160 / 9,8

R = 25 * √16,326

R = 25 * 4.04

R = 101,02 m

Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m

5 0
3 years ago
Which option describes the direction of thermal energy transfer in Earth's
Montano1993 [528]

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Explanation:

Option B..... ...

4 0
3 years ago
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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
How do you write a hypothesis
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It's and if, then statement!
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