Explanation:
Given parameters:
Displacement = 80m
Unknown:
Displacement eastward = ?
Solution:
Displacement is the distance traveled in a specific direction.
To find the distance traveled eastward, we need to find the horizontal component of his displacement using the cosine rule:
Cos Ф = 
The unknown is the adjacent;
Adjacent = hypotenuse CosФ = 80 cos 45 = 56.57m
His displacement is 56.57m due east
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To answer this you multiply the N (which is 300 in this case) by m (or 5). This will give you 1500
4) B) Speed = Distance/ time
5) C) By applying second law of newton " F = ma" we say that "15 = 12 × acceleration"
15/12= acceleration
acceleration = 1.25 m/s2
6) D) It will continue to accelerate to the right but at a slower rate because of the small force (F2) being applied from the right.
Answer:
0.71 m/s
Explanation:
We find the time it takes the stone to hit the water.
Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.
So, substituting the values of the variables into the equation, we have
y = ut - 1/2gt²
82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²
82.2 m = 0 + (4.9 m/s²)t²
82.2 m = (4.9 m/s²)t²
t² = 82.2 m/4.9 m/s²
t² = 16.78 s²
t = √16.78 s²
t = 4.1 s
This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.
Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s
Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.
Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.
The z-score is calculated by the following:

where
x is the data point
μ is population mean
σ is standard deviation
For the <u>northern</u> <u>region</u> birds:
μ = 10, σ = 3, x = 13

z = 1
The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.
For the southern region:
μ = 16, σ = 2.5, x = 13

z = -1.2
The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.