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2 years ago

Propose a hypothesis for how the position of the ball will affect the amount of its gravitational pull energy

1 answer:

4 0

<em><u>throwing a ball up initially has a lot of kinetic energy because it is moving upwards ( kinetic energy is energy which a body possesses by virtue of being in motion.) this all then get converted to gravitational potential energy, and for a moment it is stationary before it begins to fall again. by the time it has returned again, all the gravitational potential energy has turned back into kinetic.</u></em>

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Moving force of air flows through areas of high pressure to areas of low pressure

vazorg [7]

The answer is true i hope this helps

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2 years ago

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Let's practice calculating the frictional force of a skier on old "woody" skis and wet snow. The skier has a mass of 58 kg. The

kicyunya [14]

Answer:

Kinetic frictional force will be equal to 56.84 N

Explanation:

We have given mass of the skier m = 58 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of kinetic friction $\mu _k=0.1$

We have to find the kinetic frictional force

Kinetic frictional force is given by

$F_K=\mu _Kmg=0.1\times 58\times 9.8=56.84N$

So kinetic frictional force will be equal to 56.84 N

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3 years ago

What is the heaviest planet in the worlx

Sever21 [200]

The heaviest planet in the world is Jupiter!

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2 years ago

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What is the speed of a light ray (f 5.09 x10 14 Hz) in corn oil?

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Answer:

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Explanation:

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2 years ago

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The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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2 years ago