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2 years ago

Is stretching before working out important?

Physics

2 answers:

Damm [24]2 years ago

8 0

Answer:

Yes, you can twist or break parts of your body if you do not stretch before working out. It is always important to stretch even if you are doing activities such as gymnastics and karate.

Korolek [52]2 years ago

4 0

Answer:

yes

Explanation:

This is because stretching helps the muscles get ready before actually working out, so you don't strain or twist something.

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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

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2 years ago

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2 years ago

A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re

forsale [732]

Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

Now,

(1) we solve for the initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

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2 years ago