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3 years ago

What is the daily use of the deltoid?

Physics

2 answers:

Ghella [55]3 years ago

6 0

Answer:

In our everyday life, the deltoids take on most of the work in rotating our arms, but besides that the deltoid muscle also allows us to carry objects at a safe distance from the body. It is also responsible from stopping a dislocation or injury to the humerus when we carry something heavy.

777dan777 [17]3 years ago

5 0

Answer:

Lifting, pushing and pulling

Explanation:

it uses it

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A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a

Mamont248 [21]

Answer:

The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

Explanation:

Given:

Mass of the driver is, $m=55\ kg$

Radius of circular turn is, $R=300\ m$

Linear speed of the car is, $v=22.2\ m/s$

Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.

The forces are:

1. Weight = $mg=55\times 9.8=539\ N$

2. Centripetal force, 'F', which is given as:

$F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N$

Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,

$\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10$ °

Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

3 0

3 years ago

2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did

Aneli [31]

Answer:

The person delivered a power of 2,558 Watt

Explanation:

Work and Power

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being the force vector and the displacement vector, the work is calculated as:

$W=\vec F\cdot \vec s$

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

$\displaystyle P=\frac {30,184}{11.8}$

P = 2,558 Watt

The person delivered a power of 2,558 Watt

4 0

3 years ago

Under what condition will kinetic friction slow a sliding object?

matrenka [14]

Explanation: Slowing Down (or Stopping) occurs when the force of kinetic friction is greater than that of the external force. This also follows Newton's first law of motion as there exists a net force on the object.

6 0

3 years ago

Define refractive index.

levacccp [35]

The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

the ratio of the velocity of light in a vacuum to its velocity in a specified medium.

8 0

2 years ago

Read 2 more answers

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

8 0

3 years ago