For this case, the first thing we are going to do is write the generic equation of motion for the vertical axis.
We have then:
Where,
- <em>g: acceleration of gravity
</em>
- <em>vo: initial speed
</em>
- <em>h0: initial height
</em>
For the first body:
For the second body:
By the time both bodies have the same height we have:
Rewriting we have:
Clearing time:
Answer:
it takes 18.31s for the two window washers to reach the same height
Answer:
Option a) circle 5 meters and 22 meters
Step-by-step explanation:
We are given the following information in the question:
A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.
a) circle 5 meters and 22 meters
b) 19 inches and 50 inches
c) 33 centimeters and 80 centimeters
Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.
Answer: The distance between the girls is 362.8 meters.
Step-by-step explanation:
So we have two triangle rectangles that have a cathetus in common, with a length of 160 meters.
The adjacent angle to this cathetus is 40° for Anna, then the opposite cathetus (the distance between Anna and the tower) can be obtained with the relationship:
Tan(A) = opposite cath/adjacent cath.
Tan(40°) = X/160m
Tan(40°)*160m = 134.3 m
Now, we can do the same thing for Veronica, but in this case the angle adjacent to the tower is 55°
So we have:
Tan(55°) = X/160m
Tan(55°)*160m = X = 228.5 m
And we know that the girls are in opposite sides of the tower, so the distance between the girls is equal to the sum of the distance between each girl and the tower, then the distance between the girls is:
Dist = 228.5m + 134.3m = 362.8m
Answer:
1.79553
Step-by-step explanation:
Step 1: Write expression
log₁₀(√(69.5² - 30.5²))
Step 2: Evaluate square root
log(√(4830.25 - 930.25))
log(√3900)
log(62.45)
Step 3: Find log
log₁₀ (or log)
log(62.45) = 1.79553
-------------------------------------
To use the table method. we find values that are easily evaluated by log₁₀
log₁₀(10) = 1
log₁₀(50) = 1.69897
log₁₀(100) = 2
So we know that log₁₀(62.45) is between 1 and 2 and greater that 1.69897.
Where is the table..? You didn't provide any pictures..
