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max2010maxim [7]
3 years ago
13

Answer these, and tell me HOW you got to the answer.

Physics
1 answer:
alex41 [277]3 years ago
4 0
Here are the answers :)

\boxed{\bf{27)~Contrast}}

\boxed{\bf{45)~The~edges~of~objects}}<span>

</span><span>\boxed{\bf{67)~Light~comes~from~above}}

</span>\boxed{\bf{78)~Eyes~and~mouth}}

\boxed{\bf{88)~All~of~the~above}}

Hope that helps you! :)
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To cite a source is to ______.
Aliun [14]

Answer:

B

Explanation:

Mark as brailiest >_<

3 0
3 years ago
Hello, why do all electromagnetic waves travel at the same speed in vacuum?
Viktor [21]

Answer:

due to the magnetic field

Explanation:

magnetic field is the same in the vacuum

6 0
3 years ago
Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
a 5-kg fish swimming at 1 m/s swallows an absent minded 1-kg fish at rest. What is the speed of the large fish immediately afyer
storchak [24]

Answer:

In first case speed of fish will be \frac{5}{6} m/s.

In the second case the speed of fish \frac{1}{6} m/s

Explanation:

Given data :-

Mass of bigger fish ( m₁ ) = 5 kg.

Mass of small fish ( m₂ ) = 1 kg.

Speed of large fish ( v₁ ) = 1 m/s

Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.

Case - 1

Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s

Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v

v = speed of bigger fish immediately after lunch.

Using the conservation of momentum.

m₁* v₁ = ( m₁ + m₂)*v

5 = 6 * v

v = \frac{5}{6}  m/s.

Case -2

Speed of small fish = 4 m/s

Momentum of bigger fish before lunch = 5 kg.m/s

Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s

Net momentum before lunch = 5 - 4 = 1 kg.m/s

Momentum of bigger fish after eating the larger fish = 6 * V

Using the conservation of momentum.

1 = 6 * V

V = \frac{1}{6} m/s.

3 0
3 years ago
When you speak into a cell phone, your sound waves must be converted into which other waveform in order to transmit a signal
JulijaS [17]
They must be converted into radio waves in order to transmit a signal.
4 0
3 years ago
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