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Tpy6a [65]
3 years ago
11

a backpack has a mass of 8 kg. it is lifted and given 54.9 J of gravitational potential energy. how high is is lifted? accelerat

ion due to gravity is g=9.8 m/s^2.
Physics
1 answer:
dalvyx [7]3 years ago
8 0
P=mgh
h=P/mg
h=(54.9)/(8*9.8)= 0.7m
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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent
anzhelika [568]

Answer:

The position is  P  =  47.4 \ m relative to the base of the ocean

Explanation:

From the question we are told that

    The angle made by the incline with the horizontal is \theta =  24.0 ^o

    The constant acceleration is a =  3.82 \ m/s^2

    The distance covered is d =  60.0 \ m

    The height of the cliff is h  =  50 .0 \ m

The velocity of the car is mathematically represented as

      v^2 = u^2 +  2ad

The initial velocity of the car is  u= 0

So

     v^2 =  2ad

substituting values

     v^2 = 2 *  3.82 * 60

    v =  21.4 \ m/s

The vertical component of this velocity is

    v_v  =  -v * sin(\theta )

substituting values

    v_v  = -21.4 * sin(24.0)

    v_v  = -8.7 \ m/s

The negative sign is because is moving in the negative direction of the y-axis

The horizontal  component of this velocity is

     v_h  =  v * cos (\theta)

    v_h  =  21.4 * cos (24.0)

    v_h  = 19.5 \ m/s

Now according to equation of motion we have

     h  =  v_v*t  - \frac{1}{2} *  g t^2

substituting  values

    50  =  -8.7 t  - \frac{1}{2} *  9.8 t^2

    4.9t^2  +8.7t -50 = 0

using quadratic equation we have that

  t_1 = 2.42\ s \ and\   t_2 =  -4.20\ s

given that time cannot be negative

      t = 2.42 \ s

The  car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

           P  =  v_h * t

substituting values

          P  =  19.5 *  2.43

         P  =  47.4 \ m

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The two elements are in the same period, with Element R the first element in the period and Element Q the last element.
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