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Sergio039 [100]
1 year ago
10

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

Physics
1 answer:
wolverine [178]1 year ago
6 0

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

\omega = Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

1000J > \frac{1}{2}mv^2

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2

1000J = \frac{1}{2}mv^2

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

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The discovery and characterization of cathode rays was important in the development of the atomic theory because
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All matter contained electrons

Explanation:

The discovery and characterization of cathode ray suggested that it was a subatomic particle and cathode ray ( electron) was the first discovered. It immensely became the strong explanatory tool for chemical bond. This can be attributed to  the the ease with which electron move from one atom to the other.  

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria
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Answer:

Approximately \rm 11.2 \; N \cdot kg^{-1} at that distance from the center of the planet.

Option A) The low value of g near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of g on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

\displaystyle \frac{G \cdot M \cdot m}{R^2},

where

  • G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2.
  • M is the mass of the planet, and
  • m is the mass of the object.

To find an equation for g, divide the equation for gravity by the mass of the object:

\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}.

In this case,

  • M = 5.68\times 10^{26}\; \rm kg, and
  • R = 5.82 \times 10^7\; \rm m.

Calculate g based on these values:

\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}.

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for g:

\displaystyle g = \frac{G \cdot M}{R^2}.

The mass of the planet is in the numerator. If two planets are of the same size, g would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since R is in the denominator of g, increasing the value of R while keeping M constant would reduce the value of g. That explains why the value of g near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately 9.81\; \rm N \cdot kg^{-1}.

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Answer:

 

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