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3 years ago

Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

1 answer:

7 0

Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2

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Write any three reasons to separate the mixtures?

dmitriy555 [2]

Answer:

people separate mixtures in order to ger a specific substance that they need. 2. the other thing people separate mixtures to remove the unrequired components. 3.To get a pure substance

Explanation:

hope this helps probably don't because you asked this a week ago but have a good day :) ❤

5 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10-12 W/m2.

ahrayia [7]

Answer:

0.0018 W/m²

Explanation:

Power and intensity are related as:

$I=\frac{P}{4\pi r^2}$

P= 20.0 W (given)

r = 30.0 m (given)

$I=\frac{20.0}{4\pi(30.0)^2}=0.0018 W/m^2$

Intensity in decibels:

$I(dB)=10log\frac{I}{I_o}\\=10log\frac{0.0018}{10^{-12}}=92.5 dB$

7 0

3 years ago

Which statement best explains the relationship between the electric force between two charged objects and the distance between t

spin [16.1K]

Unfortunately, the given statements are missing from the problem. However, we can still determine the relationship between the electric force between two objects and the distance between them. The formula for the electric force is given below:

F = (k*Q1*Q2)/d^2

k is a constant, while Q1 and Q2 are the respective charges of the objects. F is force, while d is distance.

As seen in the formula, we can see that the electric force F is inversely proportional to the square of the distance between the two objects.

3 0

3 years ago

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Natalka [10]

By using the combined gas law which says that P1V1/T1 = P2V2/T2, assuming constant pressure, the volume at the required temperature can be obtained.((0.002 m3)(50+273.15))/(20+273.15) = volume 2 at 50 degrees C.
This gives the answer 0.0022048 m3

5 0

3 years ago

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