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3 years ago

If you are doing a "rowing" motion, what muscle of the shoulder are you using?

1 answer:

6 0

The upper back muscles being worked while using a rowing machine .your upper trapezius and rhomboids located between your shoulder blades, and latissimus dorsi located beneath the armpits

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A train travels 8.81 m/s in a -51.0° direction.

Amiraneli [1.4K]

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

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8 0

1 year ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago

A ruby laser delivers a 16.0-ns pulse of 4.20-MW average power. If the photons have a wavelength of 694.3 nm, how many are conta

stepan [7]

Answer:

The value is $n = 2.347 *10^{17} \ photons$

Explanation:

From the question we are told that

The amount of power delivered is $P = 4.20 \ M W = 4.20 *10^{6} \ W$

The time taken is $t = 16.0ns = 16.0 *10^{-9} \ s$

The wavelength is $\lambda = 694.3 \ nm = 694.3 *10^{-9} \ m$

Generally the energy delivered is mathematically represented as

$E = P * t = \frac{n * h * c }{\lambda }$

Where $h$ is the Planck's constant with value $h = 6.262 *10^{-34} \ J \cdot s$

c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$4.20 *10^{6} * 16*10^{-9}= \frac{n * 6.626 *10^{-34} * 3.0*10^{8} }{694.3 *10^{-9}}$

=> $n = 2.347 *10^{17} \ photons$

4 0

3 years ago

The distiction between long and short term memory

RUDIKE [14]

Answer:

The main difference between short term and long term memory is that the short term memory stores data temporarily while the long term memory stores data permanently. Moreover, the short term memory is volatile while the long term memory is nonvolatile. Memory is the component in a computer that stores data and information.

Explanation:

4 0

3 years ago

a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte

Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

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5 0

1 year ago