The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)
[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²
Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N
Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
Answer:
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
That is z with a pvalue of 
, so Z = 1.645.
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.09 - 3.25 = 30.84.
The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Answer:
B
Step-by-step explanation:
Hmmm this is a little hard I believe c
Answer:
600
Step-by-step explanation:
Given,
Numbers are 825 and 213.
If the number are equal or greater than 50 then one number will be rounded up. if the number is less then it will rounded down.
Like in this case 25 and 13 are less than 50 so, they will not rounded up.
825 will be rounded to 800.
213 will be rounded to 200.
Subtraction of number
= 800 - 200 = 600
