A hummingbird flies 2.0 m along a straight path at a height of 5.3 m above the ground. Upon spotting a flower below, the humming
bird drops directly downward 2.4 m to hover in front of the flower. a) What is the magnitude of the hummingbird’s total displacement? Answer in units of m.
1 answer:
To solve this problem we will apply the concepts of Pythagoras to find the net path. An easy way is to graph the path to trace the path, which resembles that of a right triangle. The information provided is equivalent to that of the two short sides of the triangle, so we will find the longest side by the Pythagorean theorem. In this way,
Therefore the magnitude of the hummingbird’s total displacement is 3.12m
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Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
<h2>Answer:</h2><h2>Explanation:</h2>
First, let's refer to the distance formula:
, where d is distance, v is velocity or speed and t is time.
Now, let's find the distance covered by each individual speed that the car had:
<h3>1. Speed 1.</h3>In order to use the formula, we need to convert minutes into hours since the speed is given in km/h.
21.1 min/60= 0.35 h.
Now, apply the distance formula.
d=(0.35h)*(86.8km/h)= 30.38 km.
<h3>2. Speed 2.</h3>Convert minutes to hours again and do the same calculations.
10.6min/60=0.18h
d=(0.18h)*(106km/h)= 19.08 km.
<h3>3. Speed 3.</h3>36.5min/60= 0.61h
d=(0.61h)*(30.9km/h)= 18.85 km.
<h3>4. Obtain the total distance.</h3>The total distance must be given by the addition of all individual distances traveled by the car on each speed:
Total distance= 30.38 km + 19.08 km + 18.85 km= 68.31 km.
Answer:
Explanation:
distance between ship A and B = 32 mile
Ship A velocity in south, dx/dt = -16 mph
Ship B is sailing toward east with speed, dy/st = 12 mph
time = 1 hour
rate of change of distance between them = ?
x be the distance travel after t time
X = 32 + x
Let distance between them be z
now, using Pythagoras theorem to calculate distance between ships after 1 hours
z² = x² + y²
z² = (32 + x)² + 12²
z² = (32 - 16)² + 12²
z = √400
z = 20 miles
now, calculation of rate of change of distnace
z² = (32 + x)² + y²
differentiating both side w.r.t. time
hence, the rate is the distance between them changing at the end of 1 hour is equal to