Answer:
0.0133A
Explanation:
Since we have two sections, for the Inductor region there would be a current
. In the case of resistance 2, it will cross a current
Defined this we proceed to obtain our equations,
For
,


For
,


The current in the entire battery is equivalent to,


Our values are,




Replacing in the current for t= 0.4m/s



From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
Learn more about power expended:brainly.com/question/11579192
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The bimetallic strip in a fire alarm is made of two metals with different expansion rates bonded together to form one piece of metal. Typically, the low-expansion side is made of a nickel-iron alloy called Invar, while the high-expansion side is an alloy of copper or nickel. The strip is electrically energized with a low-voltage current. When the strip is heated by fire, the high-expansion side bends the strip toward an electrical contact. When the strip touches that contact, it completes a circuit that triggers the alarm to sound. The width of the gap between the contacts determines the temperature that will set off the alarm.
Answer:
Force=7.71KN in the opposite direction
Explanation:
Given the mass of the car(M)=1000Kg
The initial speed of the car(u)=100km/hr
we know that 1km/hr=5/18 m/sec
100km/hr=100*5/18 m/sec=27.77m/sec
the distance travelled before it stops (s)=50m
let the acceleration of the car be a
The final velocity of the car is 0.
we know that 

a=7.712m/
We know that F=Ma
F=1000* 7.712=7712N=7.71KN