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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.

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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Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

Nady [450]

Answer:

a) W_total = 8240 J , b) W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

F - W = m a

W = mg

F = m (a + g)

F = 80 (1 + 9.8)

F = 864 N

The work of this force we will call it W1

We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

W₁ / W₂ = F y / F₂ y

W₁ / W₂ = 864/784

W₁ / W₂ = 1.1

7 0

1 year ago

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

bagirrra123 [75]

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

3 0

2 years ago

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

To eight significant figures, Avogadro's constant is 6.0221367×10^(23)mol−1. Which of the following choices demonstrates correct

Zepler [3.9K]

Answer with Explanation:

We are given Avogadro's constant = $6.0221367\times 10^{23}mol^{-1}$

There are eight significant figures.

We have to round off.

1.If we round off to four significant figures

The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth place remains same and digits on right side of ten thousandth place and ten thousandth place replace by zero.

Then ,Avogadro's constant can be written as

$6.022\times 10^{23}mol^{-1}$

If we round off to 2 significant figures

Hundredth place of given number is less than 5 therefore, digits on left side of hundredth place remains same and digits on right side of hundredth place and hundredth place replace by zero.

Then,Avogadro's constant can be written as

$6.0\times 10^{23}mol^{-1}$

If we round off six significant figures

6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.

Then, the Avogadro's constant can be written as

$6.02214\times 10^{23}mol^{-1}$

3 0

1 year ago

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2

Ostrovityanka [42]

For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then
$d ^ 2 = 16 ^ 2 + 12 ^ 2
$
From here, we clear the value of d.
We have then:
$d = \sqrt{16 ^ 2 + 12 ^ 2} 
$
Rewriting:
$d = \sqrt{256 + 144}$
$d = \sqrt{400}$
$d = 20 miles
$
Answer:
The magnitude of the car's displacement is:
d = 20 miles

7 0

1 year ago

Read 2 more answers