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Gelneren [198K]
2 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]2 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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If you peel two strips of transparent tape off the same roll and immediately let them hang near each other, they will repel each
g100num [7]

Answer:

1.

Firstly removing off one strip and it leaves electrons behind, so the strip becomes positively charged.

2. The roll however is not negatively charged because it is "earthed " by the hand holding it, thus excess negatives repel each other away through the hand.

3.Tearing off the next strip and once more it leaves electrons behind, the new strip is also positively charged and will repel the first strip.

4. Then, tear two strips apart and one will leave electrons behind on the other. Meaning that one strip is positive and the other is negative and they will attract each other.

5 0
2 years ago
Water floats on a liquid called "carbon tetrachloride." The two liquids do not mix. A light ray passing from water into carbon t
Oduvanchick [21]

Answer:

1.26

Explanation:

index  \: of  \: refraction \:  =  \:  \ \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }

\frac{ \sin(45.0) }{ \sin(40.1) }

= 1.26

5 0
2 years ago
Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer
kodGreya [7K]

The peak wavelength of Betelgeuse is 828 nm

Explanation:

The relationship between surface temperature and peak wavelength of a star is given by Wien's displacement law:

\lambda=\frac{b}{T}

where

\lambda is the peak wavelength

T is the surface temperature

b=2.898\cdot 10^{-3} m\cdot K is Wien's constant

For Betelgeuse, the surface temperature is approximately

T = 3500 K

Therefore, its peak wavelength is:

\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm

Learn more about wavelength:

brainly.com/question/5354733

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#LearnwithBrainly

8 0
2 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

8 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
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