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exis [7]
4 years ago
12

8. What happens to force when mass is increased?

Physics
1 answer:
inn [45]4 years ago
4 0

Answer:

in also increases

Explanation:

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How can you increase the intensity of sound waves
VladimirAG [237]

Answer

The intensity of a sound wave depends on the pressure of the wave,density of the medium and speed of sound in the medium. Higher density and higher sound speed both give a lower intensity. and may be it is because that sound wave is more characterize by wavelength than frequency..explanation

Explanation:

As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times greater. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.

5 0
3 years ago
You are designing a new home for a cold climate. You want one room in the house to be warmed by the Sun's energy. Which material
solniwko [45]

Answer:

glass if for a good view, but for absorbing heat concrete and brick is the best

Explanation:

wood is not a good idea bc it can cause fires from too much heat and it absorbs less heat but all of that depends on how good the material your using

8 0
3 years ago
Read 2 more answers
Compare and contrast electric potential energy and electric potential difference? Explain.
forsale [732]

Answer:

<u><em>Electric Potential Energy:</em></u>

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

<u><em>Electric Potential Difference:</em></u>

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

<u><em>Relation:</em></u>

Relation between Electric potential and electrical potential energy is given by

\delta V=\frac{PE}{q}

Here PE represents Electric potential energy

and \delta V is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0
3 years ago
A power cycle operates between hot and cold reservoirs at 1200 K and 300 K, respectively. At steady state the cycle develops a p
GenaCL600 [577]

Answer:

Explanation:

a ) Thermal efficiency = work output / heat input

= .38 MW / 1 MW = .38

OR 38%

Heat rejected at cold reservoir = heat input - work output

1 MW - .38 MW

= 0.62 MW.

b ) For reversible power output

efficiency = T₂ - T₁ / T₂   ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

= 1200 - 300 / 1200 = 900 / 1200

= .75

or 75%

rate at which heat is rejected

= 1 - .75 x 1

= .25 MW .

7 0
3 years ago
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
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