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Gelneren [198K]
3 years ago
15

The total mass of the arm shown in is 2.6 kg. Determine the force, F_M required of the "deltoid" muscle to hold up the outstretc

hed arm. Express your answer to two significant figures and include the appropriate units. Determine the magnitude of the force F_J exerted by the shoulder joint on the upper arm. Express your answer to two significant figures and include the appropriate units. Determine the angle between the positive x axis and the force F_J, measured clockwise. Express your answer using two significant figures.
Physics
1 answer:
guapka [62]3 years ago
6 0

Answer:

Fm = 51N and Fj = 26N

Summing the moments about the shoulder joint

Sum of anticlockwise moments = sum of clockwise moments

Fm x 12 = mg x 24

Fm = 2.6 x 9.8 × 24/12

Fm = 51N

Summing the forces acting on the arm

Sum of upward forces = sum of downward forces

Fm = Fj + mg

51 = Fj + 2.6 × 9.8

51 = Fj + 25.48

Fj = 51 - 25.48

Fj = 26N

Explanation:

Newtons first law and the principle of moments have been applied in solving this problem.

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I NEED HELP ASAP WILL MARK BRAINLIEST:
Evgen [1.6K]

Answer:

It is b for sure.

Explanation:

because they are examining urine now that's pure science.

7 0
3 years ago
Explain why air masses do not mix
kati45 [8]

 

Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force that preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.

If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.

At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a "front". In the military jargon, a front is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.

3 0
3 years ago
Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
3 years ago
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in
andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

7 0
3 years ago
You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
2 years ago
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