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alex41 [277]
3 years ago
9

The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in

the worst weather and road conditions. D. none of the above
Physics
2 answers:
kumpel [21]3 years ago
5 0

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

Flura [38]3 years ago
5 0

Answer:

in ideal weather and road conditions.

Explanation:

Minimum Safe Following Distances

Leave plenty of space between you and the vehicle ahead, including bicycles. If it stops quickly, you will need time to see the danger and stop.

Using the Two-Second Rule

At any speed, you can use the two-second rule to see if you are far enough behind the car in front of you:

Watch the vehicle ahead pass some fixed point - an overpass, sign, fence corner, or other marker.

Count off the seconds it takes you to reach the same spot in the road ("one thousand and one, one thousand and two...").

If you reach the mark before you finish counting, you are following too closely. Slow down and check your following distance again.  

The two-second rule applies to any speed in ideal weather and road conditions. If road or weather conditions are not good, double your following distance. You should also double your following distance when driving a motor home or towing a trailer.

Following Distance For Trucks

A truck or any vehicle towing another vehicle may not follow within 300 feet of another truck or vehicle towing a vehicle. This law does not apply to overtaking and passing, and it does not apply within cities or towns.

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Jan first uses a Michelson interferometer with the 606 nmnm light from a krypton-86 lamp. He displaces the movable mirror away f
gogolik [260]

Answer:

a) d₁ = 247.8 μm

d₂ = 205.3 μm

b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm

Explanation:

a)

The formula for Michelson Interferometer is derived to be:

d = mλ/2

where,

d = distance moved

m = no. of fringes

λ = wavelength of light

For JAN, we have following data

d = d₁

m = 818

λ = 606 nm = 606 x 10⁻⁹ m

Therefore,

d₁ = (818)(606 x 10⁻⁹ m)/2

<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>

For LINDA, we have following data

d = d₂

m = 818

λ = 502 nm = 502 x 10⁻⁹ m

Therefore,

d₂ = (818)(502 x 10⁻⁹ m)/2

<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>

b)

The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,

Resultant Displacement = Δd = d₁ - d₂

Δd = 247.8 μm - 205.3 μm

<u>Δd = 42.5 μm (in the direction of JAN)</u>

4 0
3 years ago
Urgent Important!!
Fynjy0 [20]

Answer:

B) they have been proven to cause drowsiness

Explanation:

B is the only answer that makes sense, as the other answers do not make sense.  All medications are scientifically tested before they can be used by someone.  Covering their bases would show negligence and is bad.

Hope this helps!

PS, this is biology or medical, not physics.

8 0
2 years ago
If the velocity of a bycycle is 10 m/s, how long does it take to cover a distance of 18 km? plzzz help me​
insens350 [35]

Answer:

1800 seconds

Explanation:

Given,

Distance ( d ) = 18 km

Velocity ( v ) = 10 m/s

To find : Time ( t ) = ?

Distance in km need to be converted to m.

1 km = 1000 m

18 km

= 18 x 1000

= 18000 m

So,

d = 18000 m

Formula : -

v = d / t

t = d / v

= 18000 / 10

t = 1800 seconds.

Therefore,

1800 seconds long, it will take to cover a distance of 18 km.

8 0
2 years ago
Read 2 more answers
Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)
stepladder [879]

Answer:

m(t)=20e^{-0.5t}

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope (t_{1/2}) = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

m(t)=m_0e^{-kt}

Where,

m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year

So, the equation is: m(t)=20e^{-kt}

At half-life, the mass is reduced to half of the initial value.

So, at t=t_{1/2},m(t)=\frac{m_0}{2}. Plug in these values and solve for 'k'. This gives,

\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5

Hence, the equation for the mass remaining is given as:

m(t)=20e^{-0.5t}

8 0
3 years ago
..................
kap26 [50]

Answer:

ummmm is this even a question

Explanation:

3 0
2 years ago
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