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nikdorinn [45]
1 year ago
9

The kinetic energy of a baseball that with a mass of 0.140kg is 7 J. What is the baseballs velocity?

Physics
1 answer:
Nuetrik [128]1 year ago
3 0

Answer:

10 m/s

Explanation:

Kinetic energy= \frac{1}{2}mv² where m is mass and v is velocity.

We have to make v the subject so we should rearrange the equation

K=\frac{1}{2}mv²

v²= \frac{K}{0.5*m} (use algebra)

v²= 7/0.5×0.140

v²=100

v=√100

v=10 m/s

We can confirm this by using the kinetic energy formula.

K=\frac{1}{2}×0.140×10²

K= 7 J

Hence it is proved that velocity is 10 m/s

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if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
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It would exert the same back right?
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A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
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Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

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b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

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4 0
3 years ago
(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i
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Answer:

Part a)

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

v = 2.18 \times 10^5 m/s

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T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

T = 8.2 \times 10^{15} s

Now the radius of the orbit of sun is given as

R = 3.00 \times 10^4 Ly

R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m

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Part a)

centripetal acceleration is given as

a_c = \omega^2 R

a_c = \frac{4\pi^2}{T^2} R

a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})

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Part b)

orbital speed is given as

v = \frac{2\pi R}{T}

v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}

v = 2.18 \times 10^5 m/s

5 0
4 years ago
Whats a streams discharge?
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5 0
4 years ago
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