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vladimir1956 [14]

2 years ago

15

A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3/2 km in 3.5 min. How fast (in m/s) is the car m

oving after this time?
(the equation that my teacher taught us is d= ((vi + vf)/2) x t (where d = distance, vi = initial velocity, vf = final velocity, t = time), so if you could use that it would be greatly appreciated.)

(please show all of your work or add a comprehensive explaination)

1 answer:

ArbitrLikvidat [17]2 years ago

7 0

Answer: no answer i dont no

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Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

2 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

rusak2 [61]

I'll bite:

-- Since the sled's mass is 'm', its weight is 'mg'.

-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is    (μk) times (mg) .

-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.

-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)

-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.

-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write

   s = (1/2 t²) (F - μk·mg) / m .

Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.

Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m

Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>

Unless I can come up with something a lot simpler, that's the answer.

To simplify and beautify, make the partial fractions out of the
2nd parentheses:
   <em> F (t/2) (F/m - μk·m)</em>

I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.

Five points,ehhh ?

4 0

2 years ago

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Help meee please fill in the blanks !!

Ratling [72]

The atomic number is the simply the number of protons in the atom. So in the first row with atomic number 2, the number of protons is 2.

If the atom has no charge, which I think you can assume for all of these, the number of electrons is equal to the number of protons. So the number of electrons is also 2.

The number of neutrons (which are the particles with no charge in the nucleus) is simply the mass number minus the atomic number i.e. 4 - 2 = 2.

The isotopic symbol is the symbol which is found on the periodic table of elements. There are 2 numbers associated which each element on the table. The larger is the mass number and the smaller is the atomic number. The atomic number or number of protons is what identifies the element. Looking at the periodic table ( https://sciencenotes.org/wp-content/uploads/2015/01/PeriodicTableOfTheElementsBW.pdf or https://simple.wikipedia.org/wiki/Periodic_table_(big) ), it can be seen that the element on the first row above with an atomic number of 2 is Helium with a symbol He. The number that is included with the name is simply the mass number which is 4 in this case, which tells us that this type of helium has 2 neutrons.

Another type (or isotope) of helium is Helium-3 which has one neutron.

Try the next row and post back if you have trouble with it

3 0

2 years ago

Which TWO statements describe behaviors of particles that are related to the

kolezko [41]

Answer: 1: (A) They allow electrons to move freely between them. 2: (C) they change their positions relative to one another.

Explanation:

7 0

2 years ago

There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in

qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

5 0

3 years ago

Read 2 more answers