When resistors are connected in series, they act like
a single resistor whose resistance is their sum.
100 ohms and 400 ohms, connected in series, look like
a single resistor of 500 ohms.
Current = (voltage) / (resistance)
= (60 volts) / (500 ohms) = 0.12 A.
________________________
<span>
Current is measured by connecting a meter in series
with an energized component. In other words, a break
is made in the circuit, the meter is connected in the break,
and the current to be measured literally flows through the meter.</span>
R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%
not enough information is given to determine the velocity of the object at time to=0.00s
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.