The current flowing through the circuit A is . I = e/r1 + r2.
<h3>What is circuit diagram?</h3>
An electrical circuit is depicted graphically in a circuit diagram. Simple pictures of the circuit's parts are used in pictorial circuit diagrams, where as standardised symbolic representations are used in schematic diagrams to depict the circuit's parts and connections.
"Your question is not complete, it seems to be missing the following information",
find the the current flowing through the resistors in circuit A;
The given parameters;
emf of each battery = e
internal resistance of circuit A = r₁
resistance of circuit = r₂
The equivalent resistance of the circuit A in series is calculated as follows;
Re = r1 + r2
emf = e
The current flowing through the circuit A is calculated as follows;
V = IR
I =V/R
I = e/r1 + r2
Thus, the current flowing through the circuit A is . I = e/r1 + r2
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Answer:
Explanation:
Given that,
The elevator slow to stop, this shows that it is decelerating and the final velocity is 0
v = 0m/s
Constant deceleration
a = -1m/s²
It is negative because it is deceleration
The distance the elevator descend is 4.5m
S=4.5m
Then, we want to find the time the elevator spent before stopping
Using the equation of motion
v = u + at
v² = u² + 2as
Where
v is final velocity
u Is initial velocity
a is the deceleration
s- is distance traveled
From here we can find the initial velocity of the elevator
v² = u² + 2as
0² = u² - 2 × 1 × 4.5
0 = u² - 9
u² = 9
u = √9
u = 3m/s
The initial speed is 3m/s
Then, to find the time taken, we can use the first equation
v = u + at
0 = 3 - 1 × t
0 = 3 - t
t = 3 seconds
The time taken before the elevator stop is 3 secs
Density = Mass / Volume
A) Density = 888/800 = 1.11g/mL > 1.0g/mL
B) Density = 7.8 / 8.7 = 0.897g/mL < 1.0g/mL
C) density = 725 / 715 = 1.01 g/mL > 1.0g/mL
D)density = 1.3/1.1= 1.18 g/mL > than 1.0 g/mL
E) Density = 18/19 = 0.95g/mL < 1.0g/mL
F) Density = 1.25/1.78 = 0.7 g/mL < 1.0g/mL
SUMMARY:
A C D have densities greater than 1.0g/mL
Solids
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