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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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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
1 year ago
An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me
Tasya [4]

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

3 0
2 years ago
A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0
Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

vsin(\alpha) + gt = 0

t = \frac{-vsin(\alpha)}{g}

So the vertical distance it travels within time t is

y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}

y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}

y = \frac{-v^2sin^2(\alpha)}{2g}

Similarly the horizontal distance it travels within time t is:

x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}

x = \frac{-v^2sin(2\alpha)}{2g}

We can pre-calcualte the quantity \frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54

So y = 57.54sin^2(\alpha)

x = 57.54sin(2\alpha)

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0
1 year ago
Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

6 0
1 year ago
A cyclotron particle accelerator (sometimes called an “atom smasher” in the popular press) is a device for accelerating charged
Iteru [2.4K]

Answer:

v=1.54\times 10^{7}}\ \textup{m/s}

Explanation:

Given:

The accelerated energy, U = 1.25 MeV = 1.25 × 10⁶ eV

we know,

1 eV = 1.6 × 10⁻¹⁹ J

thus,

1.25 eV = (1.6 × 10⁻¹⁹) × (1.25) J = 2 × 10⁻¹³ J

Now, Applying the law of conservation of energy, the energy due to acceleration will be equal to the kinetic energy

mathematically,

K.E = U

\frac{1}{2}mv^2=2\times 10^{-13} \ \textup{J}

where,

m = mass of the particle = 1.67 × 10⁻²⁷ kg

v = velocity of the particle

on substituting the values we get

\frac{1}{2}\times 1.67\times 10^{-27}\times v^2=2\times 10^{-13} \ \textup{J}

or

v^2=\frac{2\times 10^{-13}}{1.67\times 10^{-27}}

or

v=\sqrt{2.39\times 10^{14}}

or

v=1.54\times 10^{7}}\ \textup{m/s}

7 0
2 years ago
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