Which is not a type of driving distraction

The graph below shows the velocity of a car over time.

jekas [21]

Answer:

D. Calculate the area under the graph.

Explanation:

The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)

You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.

Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.

3 0

3 years ago

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A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the

sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0

3 years ago

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A person jogs eight complete laps around a 400-m track in a total time of 14.2 min . calculate the average velocity, in m/s.

White raven [17]

Meters per second, and the time is in minutes, so 14.2 times 60(how many seconds in 1 minute) is 852 seconds then velocity is meters divided by seconds, so 400 divided by 852 is 0.47 m/s rounded to the hundredths place

6 0

3 years ago

Which best describes the result when the concentration of a solution is increased? The amount of solute is increased, and the co

oee [108]

The amount of solute is increased, and the conductivity of the solution is increased

Explanation:

The ability of a solution to conduct electricity is conductivity.The more concentrated a solution is the higher the conductivity.This is because the concentration of ions will increase.Increasing solute increases the concentration of a solution thus rising the conductivity.

Learn More

Conductivity of a solution :brainly.com/question/9486572

Keywords : concentration, solution, solute, conductivity

#LearnwithBrainly

8 0

3 years ago

A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel

Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0

3 years ago