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Ivanshal [37]
3 years ago
14

What is the insertion for the brachialis

Physics
1 answer:
Alona [7]3 years ago
6 0

Coronoid process of the ulna

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A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12
bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction  

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair  

μ_s=F_{max}/F_{n}  

The F_n is equal to the weight multiplied by its gravity

∴F_{n}=mg  

Thus the coefficient of static friction changes as

μ_s=F_{max}/mg

μ_{s} = =165N/((25kg)\times(9.80 m/s^2 ) )

= 0.67

3 0
3 years ago
NaOH + FeCl3* Na Cl + Fe 10H)3<br> balanced
zvonat [6]
3NaOH + FeCl3 → 3NaCl + Fe(OH)3
8 0
3 years ago
Simon is riding a bike at 12 km/h away from his friend Keesha.He throws a ball at 5 km/h back to Keesha, who is standing still o
Tems11 [23]

I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.

In fact !  You didn't even say which frame the '12 km/hr' of his
bike is measured in.

So there are several different ways this could go.  I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.

-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.

-- He throws a ball at Keesha, at 5 km/hr relative to his own face.

-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.

5 0
3 years ago
What will be the final velocity of a rock if we drop it off of a bridge and it strikes the ground 2.8s later (ignoring air resis
mars1129 [50]
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time

Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s

The acceleration of the Earth when dropping something would be 9.8 m/s

Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)

5 0
3 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
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