Answer:
The time period of the empty car will be "1.00 s".
Explanation:
The given values in the question will be:
Mass,
m = 250 kg
Loaded car's time period will be:
T = 1.08 s
Shock absorbers compression,
x = 4 cm
or,
= 0.04 m
Now,
Weight of passengers will be:
⇒
The spring constant of shock absorbers will be:
⇒
As we know,
Time period,
On substituting the values, we get
(Total mass of car as well as its passengers)
Now,
The mass of the empty car will be:
⇒
hence,
The time period of empty car will be:
⇒
or,
Answer:
we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend
Explanation:
The reason for this process occurs because as we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend if they were rising, they begin to fall and if they were falling they begin to rise. Therefore the maximum point is a point of inflection of the curve since its trend changes.
Another way of looking at this process is that mathematically the point where there is a maximum or a minimum corresponds to the point where the first derivative is equal to zero, this is the slope of the line is horizontal, so the points before after correspond to values with slope of different sign.
Answer:
B) 0.0456
Explanation:
It is given that :
Rationale :
UCL = 480
LCL = 480
∴ Mean ,
The standard deviation of sample (Sn) = 11.55
Z (for UCL)
= 2
Similarly,
Z (for LCL)
= -2
Now using the z table for finding the confidence level between Z value of -2 and 2.
Confidence level = 0.4772 + 0.4772
= 0.9544
Risk (alpha) = 1 - confidence level
= 1 - 0.9544
= 0.0456
Answer:
number 1
Explanation:
they have common ancestors
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm