E L asked in Science & MathematicsPhysics · 1 month ago

May I please get help with physics?

A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.

1. What is the initial horizontal velocity, vxi, in m/s?

2. What is the initial vertical velocity, vyi, in m/s?  (+ is for up, - is for down)

3. What is the horizontal acceleration, ax, in m/s2?

 22. The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km. A 750 kg satellite orbits 35,800 km above Earth in a perfectly circular orbit.  What is the gravitational force on the satellite due to Earth?

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  • 1 month ago
    Favourite answer

    (1) The initial horizontal velocity Vxi,is given as 40 m/s.

    (2) The initial vertical velocity Vyi is given as 0 m/s

    (3) Once the ball leaves the hand the only force is gravity in the vertical direction, thus the horizontal acceleration is zero.

    (22)

    F = GMm / (r + h)²

    M = 5.972 * 10^24 kg

    m = 750 kg

    r = 6,371 km

    h = 35,800 km

    G = 6.67408 × 10^-11 N•m²/kg²

    F = (6.67408 × 10^-11)(5.972 * 10^24)(750) / (6,371,000 + 35,800,000)²

    F = 168 N

  • oubaas
    Lv 7
    1 month ago

    A softball coach throws a 1 kg ball horizontally to her player with an initial velocity of 40 m/s. The ball leaves her hand at 6 m high, and her catcher is 20 meters away.

    1. What is the initial horizontal velocity, vxi, in m/s?

    Vxi = 40 m/sec

    2. What is the initial vertical velocity, vyi, in m/s? (+ is for up, - is for down)

    Vyi = zero m/sec 

    3. What is the horizontal acceleration, ax, in m/s2?

    ax = 0 m/sec^2, since no other force but gravity act on the ball

     

     22.

    The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km. A 750 kg satellite orbits 35,800 km above Earth in a perfectly circular orbit. What is the gravitational force on the satellite due to Earth?

    g = Me*G/(re+h)^2 

    g = 5.972*10^24*6.674*10^-11/(6.371+35.8)^2*10^12) = 0.2241 m/sec^2

    Fg = m*g = 0.2241*750 = 168.1 N ...(168 with just 3 significant figures) 

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