All categories

3 years ago

13. What bone runs parallel to the fibula? A. Radius B. Femur C. Tibia D. Ulna

Physics

2 answers:

Paraphin [41]3 years ago

8 0

Answer:

C, Tibia.

The tibia is located down in the lower leg. Heres what it looks like.

i hope this helped at all.

BigorU [14]3 years ago

3 0

Answer:

C. Tibia :)

Explanation:

The fibula is the long, thin bone of the lower leg. It runs parallel to the tibia (aka shin bone) and plays a role in stabilizing the ankle and supporting the muscles of the lower leg

You might be interested in

Identify each of the quantities below as either vector or scalar

natta225 [31]

Answer:

2nd answer 3.2m/s is a scalar l, all others are vectors.

Explanation:

In order to be considered as a vector a certain quantity must have both a magnitude and a direction.

The most common example is velocity. It has both magnitude and direction, like the 1st and the last answers.

The 3rd answer is an acceleration. It is also given with its direction. So we have to consider it also as a vector.

But 2nd answer has no direction mentioned, so it is considered as a scalar.

7 0

3 years ago

What will the weather be like in your area and your friend’s area over the next 24 hours?

stealth61 [152]

Answer:

hot I checked already in my phone

6 0

3 years ago

a uniform beam 6.0 metre long and weighing 4kg rests on support at p and q placed left and right 1.0 metres from each end of the

QveST [7]

Answer:

Explanation:

I need the answer me too

8 0

3 years ago

Read 2 more answers

A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and

MariettaO [177]

Answer:

$f=140\ N$

Explanation:

Given:

mass of the object on a horizontal surface, $m=50\ kg$

coefficient of static friction, $\mu_s=0.3$

coefficient of kinetic friction, $\mu_k=0.2$

horizontal force on the object, $F=140\ N$

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

$F_s=\mu_s.N$

where:

$N=$ normal force of reaction acting on the body= weight of the body

$F_s=0.3\times (50\times 9.8)$

$F_s=147\ N$

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

$f=140\ N$

8 0

3 years ago

Brainliest for whoever is first and right.

devlian [24]

The answer is 9.8, did this last year in AP Science

3 0

3 years ago