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3 years ago

Which of the following requires constant agonist-antagonist muscle contraction

Physics

1 answer:

vodka [1.7K]3 years ago

5 0

Answer: Dynamic balance

Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121

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A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

5 0

3 years ago

Which of these are scientific questions? Check all that apply. How will climate change affect forests? Which planet is the prett

Naddika [18.5K]

Answer:

The scientific questions here are:

a) How will climate change affect forests

b) How did life on Earth begin

c) Why did dinosaurs go extinct

Explanation:

Scientific question are logical quantifiable questions, whose answers can be measured. A good scientific question must have answers that can be tested by a carefully designed experiment or measurement. Some qualities like "prettiest" and "amazing" cannot be tested for nor are they measurable, and hence, they do not make a testable component of good scientific question.

6 0

3 years ago

Please help I'm stuck on this question

Afina-wow [57]

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

8 0

2 years ago

A van moves with a constant speed of 20 miles per hour. How far can it travel in 3 1/2 hours

koban [17]

I think the answer would be 70

5 0

3 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$