(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is she going w

Question

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is she going when she reaches the bottom?

in progress 0
Bình An 3 years 2021-08-28T06:32:21+00:00 1 Answers 303 views 0

Answers ( )

    0
    2021-08-28T06:33:50+00:00

    Answer:

    v = 45.37 m/s

    Explanation:

    Given,

    angle of inclination = 8.0°

    Vertical height, H  = 105 m

    Initial K.E. = 0 J

    Initial P.E. = m g H

    Final PE = 0 J

    Final KE =  \dfrac{1}{2}mv^2

    Using Conservation of energy

     KE_i + PE_i + KE_f + PE_f

     0 + m g H = \dfrac{1}{2}mv^2 + 0

    v = \sqrt{2gH}

    v = \sqrt{2\times 9.8 \times 105}

    v = 45.37 m/s

    Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )