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Lemur [1.5K]
4 years ago
13

A good baseball pitcher can throw a ball at 100 miles/hour (about 45 m/s). If the pitcher were on Sinope, one of Jupiter’s small

er moons, could the pitcher throw the ball fast enough to escape Sinope gravity? Sinope is a roughly spherical with a radious 18000 m and mass of 10x1016 kg.
Physics
1 answer:
Zielflug [23.3K]4 years ago
4 0

Answer:

mass of sinope is 6 * 10¹⁶ kg

the ball pitcher can throw the ball fast enough to escape Sinope's gravity

Explanation:

mass of sinope m = 6 * 10¹⁶ kg

radius  r = 18000 m

the escape velocity  v = \sqrt{\frac{2GM}{r} }

v = \sqrt{\frac{2 * 6.67 * 10^-^1^1 * 6 * 10^1^6}{18000} }

v = 21.1 m/s

v is less than 45 m/s

the ball pitcher can throw the ball fast enough to escape Sinope's gravity

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TP-10 Which of the following is considered a safe refueling practice?
erik [133]

Answer:

The correct option is;

Sending all passengers below while refueling

Explanation:

Safety tips for refueling a boat includes;

1) People that are not involved in the refueling of the boat should be made to leave the area

2) The engines, electronics, ignition system and open flames should be off

3) Ensure the boat is properly secured to the dock

4) Do not use the hands-free clips and try not to be distracted

5) Ensure not to completely fill the fuel tank as fuel has the tendency to expand as its temperature rises. The tank should have about 90% filling

7 0
3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
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Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0
4 years ago
| A 1.0 kg stone is dropped from a bridge 100 m above a canyon. What will be the kinetic energy of the stone after it
Mnenie [13.5K]

Answer:

Option D

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Explanation:

When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.

Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then

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At 50 m, PE will be 1*9.81*50=490.5 J

Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence

981-490.5= 490.5 J

Approximately, 490 J

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The landing gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is d
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