Answer:
So the specific heat of the liquid B is greater than that of A.
Explanation:
Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

where:
m = mass of the body
c = specific heat of the body
change in temperature of the body
The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.
So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.


So the specific heat of the liquid B is greater than that of A.
Answer:
E = 13.2 kWh
, Cost = $ 10.8
Explanation:
We can look for the consumed energy from the expression of the power
P = W / t
The work is equal to the variation of the kinetic energy, for which
P = E / t
E = P t
look for the energy consumed in one day and multiply by the days of the month in the month
E = 110 4 30
E = 13200 W h
E = 13.2 kWh
the cost of this energy is
Cost = 0.9 12
Cost = $ 10.8
Answer with Explanation:
We are given that
Magnetic field,B=

Length of wire,l=15 m
Current,I=19 A
a.We have to find the magnitude of magnetic force and direction of magnetic force.
Magnetic force,F=
Using the formula


Direction=

15 degree above the horizontal in the northward direction.
3 bulbs are in series and if the same 3 bulbs are in parallel with the same battery then the bulbs that are connected in parallel will be dimmer
<h3>What is power?</h3>
The rate of doing work is known as power. The Si unit of power is the watt.
Power =work/time
The mathematical expression for the electric power is as follows
P = VI
The same current flows through both bulbs when they are connected in series. A greater voltage drop across the bulb with the higher resistance will result in higher power dissipation and brightness. In the case of the parallel combination, the bulb will be dimmer
Thus, If the same three bulbs are connected in series and parallel with the same battery, the parallelly connected bulbs will be dimmer, therefore the correct option is A
Learn more about power from here
brainly.com/question/3854047
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Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V