Answer:
(d) Negative.
Explanation:
let's test each at a time.
(a) It can't be 0, because cup would slide back other wise.
(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.
(c) Equal to non-conservative work done by the car's engine.
well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.
(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.
(d) this can't be true.
So the answer is (d) negative.
The sprinter’s average acceleration is 1.98 m/s²
The given parameters;
- initial velocity of the sprinter, u = 18 km/h
- final velocity of the sprinter, v = 27 km/h
- time of motion of the sprinter, t = 3.5 x 10⁻⁴ h
Convert the velocity of the sprinter to m/s;

The time of motion is seconds;

The sprinter’s average acceleration is calculated as follows;

Thus, the sprinter’s average acceleration is 1.98 m/s²
Learn more here:brainly.com/question/17280180
acceleration = Velocity changes ÷ time of the velocity changes
4 m/s^2 =
4 × 10^(-3) × 3600 km / h =
4 × 3.6 =
14.4 km / h
Thus :
14.4 = V(2) - V(1) / t(2) - t(1)
14.4 = V(2) - 20 / 10
Multiply both sides by 10
10 × 14.4 = 10 × ( V(2) - 20 ) / 10
144 = V(2) - 20
Add both sides 20
144 + 20 = V(2) - 20 + 20
V(2) = 164 Km/h
Thus the final velocity after 10 seconds is 164 Km/h .
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>.