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Artyom0805 [142]
3 years ago
9

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog on it's path 50m

ahead. If the total force applied on the breaking system on the car by the driver is 2000N what would most probably happen
Physics
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

Driver hits the dog

Explanation:

From kinematics equation

v_f^{2}=v_i^{2}+2as and making a the subject

a=\frac {v_f^{2}-v_i^{2}}{2s}=\frac {0-20^{2}}{2*50}=-4m/s^{2}

F=ma=800*-4=-3200  N

The driver is required to apply a force of 3200 N to effectively not hit the dog. However, when the driver applies 2000 N, this is less the required force by 1200 N hence the driver hits the dog.

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Answer: Is three times the height of Mt. Everest

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming
Monica [59]

Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        KE_B = \frac{1}{2} m v^2

and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

4 0
3 years ago
A lens with f = +11cm is paired with a lens with f = −25cm. What is the focal length of the combination?
Zepler [3.9K]

Answer:

19.642 cm

Explanation:

f₁ = Focal length of first lens = 11 cm

f₂ = Focal length of second lens = -25 cm

Combined focal length formula

\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm

Combined focal length is 19.642 cm

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tatiyna
Yes that would be correct. But if its an essay I'd be a bit more elaborate and give examples.
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