Answer:
Explanation:
The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark.
If the resistor in the circuit is 1 megaohlm, the total resistance in daylight and darkness will be 1.01 megaohms and 1.1 megaohlms.
The percentage difference = (1.1-1.01)/1.1*100% = 8.18%
If the resistor in the circuit is 25 kohlm, the total resistance in daylight and darkness will be 35 kohms and 125 kohlms.
The percentage difference = (125-35)/125*100% = 72%
With the input p.d to the sensing circuit fixed at 12 v, the sensing current will change according to the total resistance. A 72% difference is much more detectable. So the 25 kohm resistor is the better choice.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The earth is the fifth largest planet in our solar system.
Answer:
Speed is greater in trial 2
Explanation:
"<em>Indicate whether the speed of the block relative to the table when the block reaches the bottom of the plane is greater in trial 1 or trial 2.</em>"
The initial potential energy of the block is converted into kinetic energy, with no work done by friction.
In Trial 1, the ramp is free to move, so both the block and the ramp have kinetic energy.
PE = KE + KE
mgh = ½ mv₁² + ½ Mv²
In Trial 2, the ramp is fixed, so only the block has kinetic energy.
PE = KE
mgh = ½ mv₂²
Setting the expressions equal:
½ mv₁² + ½ Mv² = ½ mv₂²
Therefore, v₂ must be greater than v₁.