Answer:
answer a: a large front gear with a small back gear
answer b: a small front gear with a large back gear
Explanation:
just simple gearing ratios
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>.
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution =
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:

Plug in the given values and solve for
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².