By definition, power is the amount of energy consumed (or produced) in a second. (or more precisely, it is the rate of change in energy).
so anything which uses energy in a known time period can be labeled with a power rating.
an example for power could be a nuclear plant; traditional nuclear plants produce somewhat close to 1 giga watts (which means 1 giga joules in a second)
Explanation:
The speed of sound wave only depends on the property of the medium like density and the bulk modulus of the medium particle. The speed of sound also depends on the temperature of the medium.
On comparing sound waves with different frequencies and wavelengths traveling through air, the speed of the wave doesn’t depend on the frequency or the wavelength. Hence, the correct option is (1).
<u>Answer</u>
48 Volts
<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;
Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage
Np/Ns = Vp/Vs
10/4 = 120/Vp
Vp = (120 × 4)/10
= 480/10
= 48 Volts
Answer:
the answer choice is B
Explanation:
as we move from left to right , the atomic size decreases due to higher number of protons in the nucleus, which are able to attract the electrons more strongly. and so the electronegativity and electron affinity increases for the same reason. the nuclear charge increase due to more protons , and without an increase in inner electrons , there is less shielding effect. so effective nuclear charge increases.
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>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 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>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;
<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>
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
