Answer:
1. Industrial revolution was initiated or borne through the production of Steel
2. World War 1 led to the development of Tanks
Explanation:
The production of Steel through the Bessemer Process in the middle of the nineteenth century was a major technological development that spurred the Industrial revolution. This invention led to the widespread use of steel in the production of many things including vehicles and airplanes.
During the First World War in 1914, soldiers found the use of just their armaments in battle as not so productive. This led to the development of Tanks in 1915 that would continue moving towards the enemy even when being shot at.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
b
Explanation:
only if there signal is turned on
Answer: A capacitor connected across the output allows the AC signal to pass through it and blocks the DC signal, thus acting as a high pass filter. The output across the capacitor is thus an unregulated filtered DC signal. This output can be used to drive electrical components like relays, motors, etc.
Explanation:
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