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koban [17]
4 years ago
12

What is the only way to let your body recover from overtraining? A. Take vitamins B. Rest for several days C. Eat a lot of prote

in D. Drink energy drinks
Physics
2 answers:
Vladimir [108]4 years ago
8 0

Answer:

Well, you would have to rest for several of days-Apex

Explanation:

Luba_88 [7]4 years ago
7 0

Answer:

B) Rest for several days

Explanation:

Apex

You might be interested in
Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri
9966 [12]

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

a= 5.625 * 10^{-4} m/s^2

8 0
4 years ago
Read 2 more answers
How much does the gravitational force of attraction change between two asteroids if the two asteroids drift three times closer t
Katen [24]

Answer:

Increase 9 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 \times 10^{-11} m^3/kgs^2 is the gravitational constant. M_1, M_2 is the masses of the 2 objects. and R is the distance between them.

Since the force is inversely proportional to the distance squared, if it is reduced by 3 times, the gravitational force between them would increase by 3^2 = 9 times

6 0
3 years ago
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
Present day glaciers are found primarily in _______________.
Mazyrski [523]
<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>
  • <em><u>The two major ice sheets that exists today are found primarily in Antarctica and Greenland. Ice sheets are large masses of glacial ice that are also known as continental glaciers.</u></em>
  • Most ice in Antarctica and Greenland spill out into the ocean from a few spots. The Antarctica and Greenland ice sheets combined comprise more than 99 percent of freshwater ice found on Earth.
8 0
3 years ago
Read 2 more answers
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