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mafiozo [28]
3 years ago
10

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

60 mph?
Physics
1 answer:
amid [387]3 years ago
5 0
Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
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A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a
gogolik [260]

Answer:

W= 638.1 J

Explanation:

As we know that

Work done is the area of the force and displacement diagram.

W=∫F.dx

W=Work

F=force

dx=Elemental displacement

From the diagram ,area A

A= 60 x 7.09 + 1/2 x 60 x 7.09

A= 638.1 J

So the work W

W= 638.1 J

3 0
3 years ago
Differentiate conductors and insulators with the help of suitable examples
andrew-mc [135]

Answer:

Any material that keeps energy such as electricity, heat, or cold from easily transferring through is an insulator

A conductor is a material which contains movable electric charges.

3 0
2 years ago
Read 2 more answers
A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi
kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well  nitrate per 15 \mathrm{m}^{3} of water

nitrate in distant reservoir =5 \mathrm{~g} / \mathrm{m}^{3}

Let the flow rate of well

flow rate of reservoir =y m^{3} / s

Drinking water requirement is 45 \mathrm{ppm} or 45 \mathrm{~g} / \mathrm{m}^{3}

therefore, the total flow of drinking water

 

8 0
2 years ago
What is the distance from the earth's center to a point outside the earth where the gravitational acceleration due to the earth
Crazy boy [7]
R2^ 2 / R1 ^2 = g1 / g2 = 38 

<span>R2 = R1 x √38 = 6.1644* R1 </span>

<span>R2 = 6.1644 x 6378 000 = 39316632.5 m</span>
8 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
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