Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle =
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 ×
× 2
applied voltage = 3 mV
so current will be
current =
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q =
.....................1
Q =
.....................2
now we compare both equation 1 and 2 and put here value
solve it and we get
Tso = 28.153113
so Tso = 28.15°C
The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.
Define work. Explain the rate of doing work.
Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.
Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.
Solution Explained:
Given,
Weight = 1000N and distance = 5m
A/Q, the work here is done in lifting then
Work = (weight) × (distance moved)
= 1000 X 5
= 5000Nm or 5000J = 5kJ
Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.
To learn more about work, use the link given
brainly.com/question/25573309
#SPJ9
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Answer:
Explanation:
First of all,we need to know what kind of research and how important it is before determining which computation method to use .
Then compare end result of research and cost of expensive methods .
Then we should opt for the best expensive method to be used for the given application.
Answer:
a fluid power engine okk done please