Low air pressure systems are usually associated with which type of weather?

After each charging, a battery is able to hold only 99% of the charge from the previous charging. The battery was used for 5 hou

Aloiza [94]

Answer:

500 hours

Explanation:

The sum of total hours over its lifetime will be given by

$T=\frac {T_o}{1-r}$

Where T is total time, r is rate in decimal and To is the original charge hours. Substituting the original charge hours with 5 hours and rate as 0.99 then the time will be

$T=\frac {5\ hours}{1-0.88}=500\ hours$

Therefore, the time is equivalent to 500 hours

7 0

3 years ago

A kayak took 5 hours to finish its trip on a river.If it traveled at an average speed of 18mph (mile per hour),What was the dist

weqwewe [10]

23feet because she went 18mph

8 0

3 years ago

Drag the item from the item bank to its corresponding match.

tigry1 [53]

1- first law
2- third law
3- first law
4- second law
5- third law
6- second law

3 0

3 years ago

Read 2 more answers

If 200 ml of tea at 90 °C is poured into a 400 g glass cup initially at 25 °C, what would be the final temperature of the mixtur

sweet-ann [11.9K]

The final temperature of the mixture is 43.62 °C

To solve the question above, we apply the law of calorimetry

The law of calorimetry: which states that if there is no lost of heat the surrounding, heat lost is equal to heat gained, or it can be stated as heat absorbed by a cold body is equal to heat released by a hot body, provided there is no lost of heat to the surrounding.

The law above is expressed mathematically as

Cm(t₁-t₃) = C'm'(t₃-t₂)............. Equation 1

Using equation 1 to solve the question,

Let: C = specific heat capacity of glass cup, m = mass of glass cup, C' = specific heat capacity of tea, m' = mass of tea, t₁ = initial temperature of tea, t₂ = initial temperature of glass cup, t₃ = final temperature of the mixture.

From the question,

Given: m = 400 g = 0.4 kg, C = 840 J/kg°C, m' = 200g (tea is a liquid made of water and the volume of water in ml is thesame a its mass in gram) = 0.2 kg, C' = 4186 J/kg.°C, t₁ = 90°C, t₂ = 25°C

Substitute these values into equation 1 and solve for t₃

0.4(840)(90-t₃) = 0.2(4186)(t₃-25)

336(90-t₃) = 837.2(t₃-25)

30240-336t₃ = 837.2t₃-20930

collect like terms

837.2t₃+336t₃ = 30240+20930

1173.2t₃ = 51170

t₃ = 51170/1173.2

t₃ = 43.62 °C

Hence, the final temperature of the mixture is 43.62 °C

7 0

3 years ago