4. Which activity would be BEST for prolonged training within your target heart rate zone?

1 answer:

6 0

I am sorry I would answer this questions but I don’t have the options of the heart rates so I can give you the best answer

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What is the gravitational force between mars and Phobos

alina1380 [7]

Answer:

$F=5.16\times 10^{15}\ N$

Explanation:

We have,

Mass of Mars is, $m_M=6.42\times 10^{23}\ kg$

Mass of its moon Phobos, $m_P=1.06\times 10^{16}\ kg$

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

$F=G\dfrac{m_Mm_P}{d^2}$

Plugging all values, we get :

$F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N$

So, the gravitational force is $5.16\times 10^{15}\ N$ .

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3 years ago

11. You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the gla

Flauer [41]

Explanation:

it happens in the morning time

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Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring

arlik [135]

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$