9. The acceleration due to gravity on the

What is the purpose of the scapula to move during arm elevation?

Inessa [10]

The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

Learn more about scapula motion at: brainly.com/question/5133017

#SPJ12

3 0

1 year ago

Which statement about elements and atoms is true?

Vika [28.1K]

Atoms are the smallest unit of an element

8 0

3 years ago

Sound travels through water at a speed of 1500 m/s. If the frequency of a sound is 1000 Hz, what is the wavelength?

ratelena [41]

Answer:

1.5m

Explanation:

Velocity=1500m/s

Frequency=1000hz

Wavelength =velocity ➗ frequency

wavelength =1500 ➗ 1000

Wavelength=1.5m

3 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

6 0

2 years ago

.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric

maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0

2 years ago