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3 years ago

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Which of the following statements is correct? Which of the following statements is correct? The more a muscle shortens, the more

power it generates. The number of muscle fibers best determines how powerful a muscle will be. Muscle fibers running in parallel arrangement generate more power. Multipennate muscles do not produce much power because the fibers run in many directions.

1 answer:

NeX [460]3 years ago

4 0

The correct answer are

a) "The number of muscle fibers best determines how powerful a muscle will be"

b) "The more a muscle shortens, the more power it generates."

Reason :

Muscle fiber in longitudinal directions generate more power

Multipennate muscles do not produce much power because the tendon branches within muscle .

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3. A motorbike travels 45 miles in 15 minutes, what is its speed?

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2 years ago

One watt is wqual to 1kg .m/s³

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A bus is traveling with a uniform acceleration of 2.75 meters/second2. If the initial velocity of the bus is 16.5 meters/second,

Kaylis [27]

<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>

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3 years ago

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l

Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

Wₐ = 1500 lb

W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

W = mg

m = W / g

mₐ = Wₐ / g

m_b = W_b / g

mₐ = 1500/32 = 46.875 slug

m_b = 125/32 = 35,156 slug

Let's reduce to the english system

v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

p_f = (mₐ + m_b) v

the moment is preserved

p₀ = p_f

mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

v = $\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}$

we substitute the values

v = $\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6$

v = 0.559 v₀ₐ - 26.40 (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

M = mₐ + m_b

M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

K₀ = ½ M v²

final point. When they stop

K_f = 0

The work is

W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

N-W = 0

N = W

the friction force has the expression

fr = μ N

we substitute

-μ W x = Kf - Ko

-μ W x = 0 - ½ (W / g) v²

v² = 2 μ g x

v = $\sqrt{ 2 \ 0.750 \ 32 \ 17.5}$ Ra (2 0.750 32 17.5

v = 28.98 ft / s

3 0

2 years ago