b. Waves can exist in liquids and solids only.
c. Waves can exist in solids and liquids only.
d. Waves can exist in solids, liquids, and gases.
Answer:
Explanation:
We should know that the gradient(slope) of a distance-time graph is the speed, and the gradient of a velocity-time graph is the acceleration.
a) Since the gradient from t=80 to 160 is 0, the speed is 0, indicating it is at rest.
b)To find the speed, we find the gradient of the graph at t=40. Since it is a straight line from t=40 to 60, we can do this simply by
Gradient = (200-120)/(60-40)=80/20=4 (This is an estimate since the graph is not clear, instead of t=40 and t=60, you can use any two points along that straight line (from about t=20 to 60))
c)The one on the right is the velocity graph, and the one on the left shows acceleration. We can know this by analysing some points. For example, and t=50, the right graph is about constant at 5, while the left one is 0. This would indicate that the left one shows the gradient of the right (gradient of a constant portion is 0). We can choose another place on the right graph, say t=60 to 80. The curve is decreasing. The left graph for this same time period is negative. This reaffirms our guess, since the gradient of a decreasing curve should be negative.
If you want to do it the other way around, you can analyse the left graph. From about t=15 to t=30 the graph is decreasing. Thus its slope should be negative. However, the right graph is positive for this period. This means that the right graph definately does not show the slope of the left.
When you learn about calculus and derivatives, these will become much clearer :)
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:
Explanation:
V=IR
I=12/(R of resistor + R of LDR)
R of LDR = 10kohm in light and = 100kohm in dark
R1 = 25kohm
R2 = 1Mohm
solve 4 current
light dark
R1 12/(25+10)=0.343mA 12/(25+100)=0.096mA
R2 12/(1000+10)=0.012mA 12/(1000+100)=0.011mA
so R1 is better as its easier 2 tell its light or dark