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Vedmedyk [2.9K]
3 years ago
7

the skid marks left by the decelerating jet-powered car the spirit of america were 9600 m long before it came to a stop at 198 m

/s, how long did it take for the car to stop ?
Physics
2 answers:
N76 [4]3 years ago
8 0

Answer:

97 s

Explanation:

Given:

Δx = 9600 m

v₀ = 198 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

9600 m = ½ (0 m/s + 198 m/s) t

t = 97 s

trapecia [35]3 years ago
6 0

Answer:

Δx = ½ (v + v₀) t

9600 m = ½ (0 m/s + 198 m/s) t

t = 97 s

Explanation:

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<span>The answer Fossil fuels aren’t widely available, but pure hydrogen is. Fossil fuels are non-renewable. Hydrogen is not a fuel. It is a way of storing or transporting energy. Distribution and infrastructure needs to be refurbished to cope with hydrogen, which can metals by making them brittle. </span>
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Answer:

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Explanation:

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n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

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