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saul85
2 months ago
8

A typical reaction time to get your foot on the brake in your car is 0.2 second. If you are traveling at a speed of 60 mph (88 f

t/s or 26.8 m/s), what distance will your car travel during this reaction time? Express your answers in miles, feet and meters.
Physics
2 answers:
Softa [3K]2 months ago
6 0

Answer:

Distance covered in meters= 5.36

Distance covered in feet= 17.6

Distance covered in miles= 3.33*10⁻³

Clarification:

The rule of three is a method for addressing proportional problems between three known values and one unknown, creating a proportionality relationship among them. The goal is to discover the fourth term of a ratio given the other three. Remember that proportionality represents a constant relationship or ratio among different quantities.

If the connection between the quantities is direct, implying that an increase in one quantity leads to an increase in the other (or, conversely, a decrease triggers a decrease), then the direct rule of three should be applied. The process for a direct rule of three follows:

a ⇒ b

c ⇒ x

Then:

x=\frac{c*b}{a}

This directs us to apply the rule of three as follows, considering a speed of 26.8 \frac{m}{s}: if you move 26.8 meters in 1 second, what distance do you cover in the reaction time of 0.2 seconds?

distance in meters=\frac{0.2 seconds*26.8 meters}{1 second}

Distance in meters= 5.36

Next, applying the same rule of three for 88 \frac{ft}{s}: if 88 feet is traveled in 1 second, how far in the reaction time of 0.2 seconds?

distance in feet=\frac{0.2seconds*88 feet}{1 second}

Distance in feet= 17.6

Finally, considering the fact that 1 hour comprises 3600 seconds, we apply the rule of three based on the speed of 60 mph: if you travel 60 miles in 3600 seconds, what's the distance covered in 0.2 seconds?

distance in miles=\frac{0.2 seconds*60miles}{3600 seconds}

Distance in miles= 3.33*10⁻³

inna [3.1K]2 months ago
5 0

Explanation:

Insufficient details. It heavily relies on the technical specifications of the vehicle (the information supplied merely pertains to reaction time in humans, without accounting for the delay involved when the brakes are applied

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A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while
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To address this problem, Boyle's Law must be applied, which states that the initial and final pressures and volumes are related as follows: Where, P₀ and V₀ represent the initial pressure and volume, while P and V refer to the final pressure and volume. The endpoint pressure in this scenario is atmospheric pressure. Thus, using the given equation, we can find the volume the lungs would occupy at the surface.
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2 months ago
The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the
kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N

In the case of the N-H-N combination:

The total force acting from this combination is:

F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N

The force that thymine applies on adenine is:

F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N

When rounded to three significant figures, the net force is 8.86\times 10^{-9}N

b)

The negative value indicates that the force is attractive, as it is aimed towards thymi.

c)

The force acting on the electron due to the proton is:

F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

d)

The ratio of the above forces is:

\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

5 0
3 months ago
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