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alexdok [17]
3 years ago
11

Which one of the following types of training involves 10 to 30 minutes of high-intensity exercise?

Physics
2 answers:
Goshia [24]3 years ago
5 0

Correct answer choice is:



C. Medium range



Explanation:



Medium range exercises are used to gain extra strength and fitness. Usually, heavyweights are used with less number of repetitions. These sort of exercises are mostly the hardest t do. All you need is to have a high level of motivation and stamina, which can be gained by running or cycling.

Evgen [1.6K]3 years ago
4 0

Answer:

c. medium range

Explanation:

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Answer: The mass would be 1000m3 and the volume would be 1000kg

Explanation:

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Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

8 0
3 years ago
A book falls off a shelf that and hits the ground at 10.0 m/s. How far did it fall?
ELEN [110]

Answer:

Explanation:

14 m/s

Explanation:

The motion of the book is a free fall motion, so it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. Therefore we can find the final velocity by using the equation:

where

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

d = 10.0 m is the distance covered by the book

Substituting data, we find

8 0
3 years ago
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