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Nimfa-mama [501]
3 years ago
12

Planet X has three times the free-fall acceleration of Earth.

Physics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

a) The ball goes one-third times higher on X

b) The ball goes three times higher on X.

Explanation:

a)

  • As the initial velocity is the same than on Earth, but the free-fall acceleration is three times larger, this means that the only net force acting on the ball (gravity) will be three times larger, so it is clear that the ball will reach to a lower height, as it will slowed down more quickly.
  • Kinematically, as we know that the speed becomes zero when the ball reaches to the maximum height, we can use the following kinematic equation:

        v_{f} ^{2} - v_{o}^{2}  = 2* \Delta h* g

       since vf = 0, solving for Δh, we have:

       \Delta h = h_{max} =\frac{v_{o} ^{2}}{2*g} (1)

       if v₀ₓ = v₀E, and gₓ = 3*gE, replacing in (1), we get:

     Δhₓ = 1/3 * ΔhE

      which confirms our intuitive reasoning.

b)    

  • Now, if the initial velocity is three times larger than the one on Earth, even the acceleration due to gravity is three times larger, we conclude that the ball will go higher than on Earth.
  • We can use the same kinematic equation as in (1) replacing Vox by 3*VoE, as follows:

       \Delta h = h_{max} =\frac{(3*v_{o}) ^{2}}{2*3*g} (2)

      Replacing the right side of (1) in (2), we get:

      Δhx = 3* ΔhE

      which confirms our intuitive reasoning also.

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An internal resistance of 5 ohm and a battery of 15 ohm is connected to a resistance of 20 ohm calculate the electric current
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Answer:

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Explanation:

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5 0
2 years ago
A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release loc
zvonat [6]

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s².  The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for <em>8.16 seconds</em> up and down.

4 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

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-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

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\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

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r = 1.32·R_E

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ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

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