8. What happens to force when mass is increased?

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:

Electric Potential Energy:

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

Electric Potential Difference:

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.

Relation:

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;

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

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

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for copper metal;

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

For aluminum metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for aluminum metal ;

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

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