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Karo-lina-s [1.5K]

3 years ago

14

If you hold your arm outstretched with palm upward, the force to keep your arm from falling comes from your deltoid muscle. assu

me that the arm has mass 4 kg and the distances and angles shown in

1 answer:

Maurinko [17]3 years ago

4 0

Your question seems to be incorrect. Please check below:

What force must the deltoid muscle provide to keep the arm in this position? By what factor does this force exceed the weight of the arm?<span>If you hold your arm outstretched with palm upward, as in (Figure 1) , the force to keep your arm from falling comes from your deltoid muscle. Assume that the arm has mass 4 kg and the distances and angles shown in (Figure 1) .

F=?
F/w= ?

The answer is </span><span>339 N</span><span>

</span>

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Answer it asap<br> I promise i will mark them the Brainly

nataly862011 [7]

Answer:

A) The acceleration is zero

<em>B) The total distance is 112 m</em>

Explanation:

<u>Velocity vs Time Graph</u>

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

$\displaystyle a=\frac{8-8}{12-0}=0$

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

Area of rectangle= base*height=12 s * 8 m/s = 96 m

Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m

The total distance is: 96 m + 16 m = 112 m

4 0

3 years ago

Which best describes a value for density

scoray [572]

Density can be any number, as long as it has the right units.

A unit of density has to be (a unit of mass) divided by (a unit of volume).
The most common one is gram/cm^3.

8 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

What happens when a light ray is parallel to the principal axis ?

Phantasy [73]

Answer:

Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.

Hope this helps...

7 0

3 years ago

A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn

olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

$v_{f} = v_{i} + a*t\\$

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0

3 years ago