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malfutka [58]
3 years ago
9

How long will it take you to pass a truck at 60 mph with oncoming traffic?

Physics
2 answers:
DedPeter [7]3 years ago
6 0
<span>This is impossible to calculate without knowing the speed of each vehicle. </span>
kobusy [5.1K]3 years ago
5 0

Answer:

4 seconds - Not practical

Explanation:

Length of the truck = 50'

Initial distance behind the truck = 30'

Finish Pass = 50' ahead of truck ,

Pass at =  60mph -- about 3.375 seconds.

- 70mph your closing speed is 130mph.  

If you were less than a 1/4 mile away when you tried the pass you will be dead.

That would be a quick pass.  You will probably want a mile beyond the oncoming traffic.

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A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?
Aleks [24]

Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

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3 years ago
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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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Describe The energy conversions that occur when coal is formed, and then burned to produce thermal energy.
svlad2 [7]

Answer:

When energy contained in coal is turned into heat, and then into electrical energy. As boiling water heated by the burning coal is cooled, steam forges from these cone-shaped cooling towers.

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I think I am right with this

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Answer:

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Explanation:

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