Answer:
The current in the circuit decrease slowly .
Explanation:
Given as :
For the electrical circuit
The voltage V in the circuit is slowly decreasing
The resistance R of the resistor slowly increasing after heating
Now, From Ohm's Law
Voltage is directly proportional to the flow of current through circuit
I.e V ∝ I
Or. V = R × I
where R is the proportionate constant and this is the resistance of the resistor
whose property is to opposes the flow of current in the circuit
So, If R value more then current I reduces in the circuit
∵ Here in the circuit , The resistance is slowly increasing, so, current I is slowly decreasing .
Hence The current in the circuit decrease slowly . answer
Answer:
3,4,5,6,7,8,9,10
3. lithium
4. beryllium
5. boron
6. carbon
7. nitrogen
8. oxygen
9. fluorine
10. neon
those are the numbers in period 2 on the periodic table.
Explanation:
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
That ratio is called"efficiency". It doesn't need to be a percent.
It can just as well be a fraction or a decimal number.
