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n200080 [17]
3 years ago
7

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 41m in front of you. Your reaction time

before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10 m/s2 .What is the maximum speed you could have and still not hit the deer?
Physics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

Maximum speed you could have and still not hit the deer = 24.07 m/s

Explanation:

Let the maximum speed you could have and still not hit the deer be y.

Your reaction time before stepping on the brakes is 0.50s.

Distance traveled during this time = 0.5y

A deer steps onto the road 41m in front of you

Remaining distance to deer = 41 - 0.5y

The maximum deceleration of your car is 10 m/s²

We have equation of motion, v² = u² + 2as

       Initial velocity, u = y m/s

       Final velocity, v = 0 m/s

       Acceleration, a = -10 m/s²

       Displacement, s = 41 - 0.5y

Substituting,

       v² = u² + 2as

        0² = y² + 2 x -10 x (41 - 0.5y)

          20(41 - 0.5y) = y²

           820 - 10 y = y²

            y² + 10 y -820 = 0

             y = 24.07 m/s or y = -34.06 m/s(not possible)

Maximum speed you could have and still not hit the deer = 24.07 m/s

           

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Answer:

Scientific method

Explanation:

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Which statement(s) correctly compare the masses of protons, neutrons, and electrons?
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Electron<span>. the central part of an atom containing </span>protons<span> and </span>neutrons<span> ... which of the following is necessary to calculate the atomic </span>mass<span> of an element? ... which of the </span>statements correctly compares<span>the relative size of an ion to its neutral atom?</span>
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3 years ago
A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

6 0
4 years ago
A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su
labwork [276]

Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

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  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

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