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3 years ago

What is the bone in your arm

2 answers:

swat323 years ago

8 0

The humerus is the only bone of the upper arm. It is a long, large bone that extends from the scapula of the shoulder to the ulna and radius of the lower arm. The proximal end of the humerus, known as the head, is a round structure that forms the ball of the ball-and-socket shoulder joint.
I just got it from google normaly i would simplify it but its to simple for me to simplify again hope this helps

Nina [5.8K]3 years ago

7 0

Here are some bones our arm have:
Humerus
Lateral epicondyle
capitulum
Head
neck
raduis
Tuberose

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Free_Kalibri [48]

Explanation:

If a person is set to walk on a constant speed regardless of the situations then if the person walks a certain distance with no interruptions in an observed time then her speed can be calculated.

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3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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3 years ago

PLS HELPPPP WILL AWARD BRAINLIEST AND POINTS

Yuri [45]

Answer:

Here

Length (l) = 6 m

height (h) = 3 m

Load(L) = 500 N

Effort (E) = ?

we know the principal that

E * l = L * h

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6E = 1500

E = 250

therefore 250 N work is done on the barrel.

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4 years ago

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3 years ago

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exis [7]

Answer:

Explanation: its weird

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