Ask question

Ask question

All categories

Arte-miy333 [17]

2 years ago

4

A 3.6-volt battery is used to operate a cell phone for an hour. If the cell phone draws a current of 0.018 Amperes during that t

ime, what is the energy lost by the battery?

2 answers:

zavuch27 [327]2 years ago

5 0

Answer:

the energy lost by the battery is 233.28Joules

Explanation:

Energy = Power × Time

We know that the power is given by :

$P=V \times I$

Power = Current (I)×potential difference (V)

Energy lost = current × potential difference × time

$E = I \times V \times t$

Given that

current, I = 0.018Amperes,

Potential diff ,V = 3.6Volts

timr, t = 1hour = 1×60×60 = 3600seconds

Energy lost. = 0.018×3.6×3600

= 233.28Joules

Hence, the energy lost by the battery is 233.28Joules

Margaret [11]2 years ago

3 0

Answer:

233.28Joules

Explanation:

Energy lost by the battery is expressed as

Energy = Power × Time

Since Power user up = Current (I)×potential difference (V)

Energy lost = current × potential difference × time

Energy used up = IVt

Given the following parameters

I = 0.018Amperes, V = 3.6Volts t = 1hour = 1×60×60 = 3600seconds

Energy lost. = 0.018×3.6×3600

Energy lost by the battery = 233.28Joules

You might be interested in

What is the mass of a large dog that weighs 441 newtons

PtichkaEL [24]

Weight, w = mg. g ≈ 9.8 m/s². m = mass in kg. w is weight in N

441 N = m* 9.8

9.8m = 441

m = 441/9.8

m = 45 kg.

Mass of the dog is = 45 kg

5 0

2 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

5 0

2 years ago

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

grigory [225]

Answer: $h=160.84 W/m^2-K$

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied $Q=mc\Delta T=0.3\times 2190\times (30)$

Heat due to convection =hA(100-30)

$A=\pi d\cdot L$

$A=\pi 0.05\times 10=1.571 m^2$

$Q_{convection}=h\times 1.571\times (100-22)=122.538 h$

$Q=Q_{convection}$

19,710=122.538 h

$h=160.84 W/m^2-K$

5 0

2 years ago

A projectile is launched at an angle of 60° from the horizontal and at a velocity of

gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

5 0

2 years ago

A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0

2 years ago