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3 years ago

There are five different types of muscles in exercise physiology.

Physics

1 answer:

9966 [12]3 years ago

3 0

Answer: The answer is B) False

Explanation: There are only three muscle types in the exercise physiology.

Skeletal Muscle

Smooth Muscle

Cardiac Muscle

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A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by

Margarita [4]

Thank you for posting your question here at brainly. Feel free to ask more questions.
The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds . 

Hope my answer would be a great help for you.

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3 years ago

S/REF No. Date If the load distance of a level is 20 cm and effort distance is 6ocm, calculate the amount of effort required to

andrew11 [14]

The answer is 602 cause I added

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3 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago

A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per seco

yanalaym [24]

Answer:

(2) 10.0 m/s east

Explanation:

7 0

3 years ago

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

Vinvika [58]

Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

The normal force which acts perpendicularly to the downhill plane
The weight of the car which acts vertically downwards
The drag force due to air resistance which acts in opposition to the motion of the car

Friction is ignored, so the force due to friction is assumed negligible

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3 years ago