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2 years ago

11

To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary

, in what instance does a person become unhealthy

1 answer:

scZoUnD [109]2 years ago

7 0

Unhealthy meals & skipping workouts

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A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh

Debora [2.8K]

Answer:

A. 5/7h

B. Same height

C. See attachment below

Explanation:

Find the attachment below for better understanding.

7 0

3 years ago

HELP!!! The planet Mars has a mass about one-tenth the mass of Earth. Even though Mars has two moons, their tidal forces have a

vladimir2022 [97]

Answer:

Say: Mars has a much weaker gravity effect than it does because it is smaller and cannot have as much gravity effect than it does on earth.

Explanation:

4 0

3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

kifflom [539]

Based on the law of conservation of energy, we know that we can't create energy, machines can only convert one type of energy into another. So, if we want to improve a machines's ability then we need to reduce it's loss energy (part of energy which is useless). Out of all the options only Option C fits best with it.

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago

Read 2 more answers

Which contains the particles that are vibrating fastest?

Phantasy [73]

Answer:

Boiling water

Explanation:

3 0

2 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for <em>x</em> :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago