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nordsb [41]
3 years ago
11

intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr

avelled​
Physics
2 answers:
V125BC [204]3 years ago
4 0

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

Usimov [2.4K]3 years ago
4 0

Answer:

Acceleration=12.5/3m/s^2=4.17m/s^2

Explanation:

Initial velocity u = 45 km/h = 12.5 m/s; Final velocity v = 0 m/s; time t = 3 s, acceleration a.

[a=v-u/t] : a=0-12.5/3=-12/3m/s^2

(Retaradation=-(-12.5/3)=12.5/3.

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[ V = I x R ] V (volts) = I (amps) x R (Ω)

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2 years ago
19. When white light is incident on a prism, which one of the resulting color components will have the lowest index of refractio
ohaa [14]

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option A

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5 0
3 years ago
A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. W
Lilit [14]

Complete Question

 The complete question is shown on the first uploaded image  

Answer:

The total pressure is  P_T  = 10.79*10^{5} N/m^2

The temperature at the bottom is T_b  = 284.2 \ K

Explanation:

From the question we are told that

    The length of the glass tube is  L  = 1.50 \ m

      The length of water  rise at the bottom of the lake  d = 1.33 \ m

     The depth of the lake is  h =  100 \ m

     The air temperature is T_a   =  27 ^oC =  27 +273 =  300 \ K

      The atmospheric pressure is  P_a = 1.01 *10^{5} N/m

      The density of water is \rho =  998 \ kg/m^3

The total pressure at the bottom of the lake is mathematically represented as

                 P_T  =  P_a + \rho g h

substituting values

               P_T  = 1.01*10^{5}  + 998 * 9.8 * 100

               P_T  = 10.79*10^{5} N/m^2

According to ideal gas law

         At the surface the glass tube not covered by water at surface

             P_a V_a =  nRT_a

Where is the volume of

             P_a *A * L  = nRT_a

 At the bottom of the lake  

           P_T V_b  =  nRT_b

Where V_b is the volume of the glass tube not covered by water at bottom

          and  T_b i the temperature at the bottom

  So the ratio between the temperature  at the surface to the temperature at the bottom is mathematically represented as

             \frac{T_b}{T_a}  =  \frac{d * P_T}{P_a * h}

substituting values

           \frac{T_b}{27}  =  \frac{0.133  * 10.79 *10^5}{1.01 *10^{5} * 1.5}

   =>     T_b  = 284.2 \ K

           

       

3 0
3 years ago
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