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sp2606 [1]
3 years ago
14

The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w

hat is the magnitude of the stabilizing component of the brachialis force
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

XY=636N

Explanation:

From the question we are told that:

Distance d=0.35m

Angle \theta=32\textdegree

Force F=750N

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

 XY=Fcos\theta

 XY=750cos 32\textdegree

 XY=636N

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Answer:

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Explanation:

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3 years ago
Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p
miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H  

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

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3 years ago
A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
jok3333 [9.3K]

Answer:

S_a_v_e_r_a_g_e=48km/h

Explanation:

Ok, the average speed can be calculate with the next equation:

S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}   (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

Total\hspace{3}distance=2*d

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

Total\hspace{3}time=t1+t2

From basic physics we know:

t=\frac{d}{S1}

so:

t1=\frac{d}{S1}

t2=\frac{d}{S2}

Using the previous information in equation (1)

S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }

Factoring:

S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}   (2)

Finally, replacing the data in (2)

S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h

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3 years ago
__ is the apparent in the position of an object when viewed from two locations
Yakvenalex [24]
The apparent shift <span>is the apparent in the position of an object when viewed from two locations.

Hope this helps :)</span>
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