Answer:
Available options are;
A. validity; high
B. variability; low
C. reliability; low
D. reactivity; high
Answer is C. reliability; low
Explanation:
Normal body temperature is considered to be 37°C (98.6°F); however, a wide variation is seen. Among normal individuals, mean daily temperature can differ by 0.5°C (0.9°F), and daily variations can be as much as 0.25 to 0.5°C. So we can see that the 98 to 106 to 89 degrees shows that the thermometer is not reliable.
Answer:
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
Explanation:
For this exercise we draw a Cartesian coordinate system in this system: East coincides with the positive part of the x-axis and North with the positive part of the y-axis.
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
In the attachment we can see the representation of the three vectors
For Part A.
The absolute value of vector A is
<span>|A | = 6
This is simply the square root of the sum of the squares of the coordinates.
For Part B
Do the same thing as in Part A
| B | = </span>√14
<span>
Part C
Simply subtract the corresponding elements resulting in
(-3, 2, 7)
Part D
</span>√62
<span>
Part E
Yes</span>
Answer:
The distance is 709.5 m.
Explanation:
Given that,
Speed = 150 m/s
Distance = 110 m
Suppose, How far short of the target should it drop the package?
We need to calculate the time
Using equation of motion


Where, g = acceleration due to gravity
t = time
Put the value into the formula


We need to calculate the distance
Using formula of distance

Put the value into the formula


Hence, The distance is 709.5 m.
Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s