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4 years ago

The pulse site located at the point where the upper leg bends is called the

1 answer:

3 0

The pulse site located at the point where the upper leg bends is called the femoral. It is an artery found in the thigh. It is large and is deemed as the main arterial supply for the lower part of the body. It is known as the second artery that is the largest. It is being used as the catheter access artery. From it, diagnostics for the heart, brain, arms, kidney and other parts can be directed to the other arterial system. It can also be used as a source to draw blood that is from the arteries when there is low blood pressure.

You might be interested in

What was the hikers average velocity during part D of the hikes?

pychu [463]

Velocity = displacement / time
v = 6 / 3/4
v = 6*4/3
v = 24/3
v = 8 m/s toward north

In short, Your Answer would be 8 m/s north

Hope this helps!

8 0

4 years ago

Read 2 more answers

A 93.5 kg snowboarder starts from rest and goes down a 60 degree slope with a 45.7 m height to a rough horizontal surface that i

frosja888 [35]

Answer:

a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m

Explanation:

This can be answered with either force analysis and kinematics, or work and energy.

a) Using force analysis, we can draw a free body diagram for the snowboarder. There are two forces: normal force perpendicular to the slope and gravity down.

Sum of the forces parallel to the slope:

∑F = ma

mg sin θ = ma

a = g sin θ

Therefore, the velocity at the bottom is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)

v = 29.9 m/s

Alternatively, using energy:

PE = KE

mgh = 1/2 mv²

v = √(2gh)

v = √(2×9.8×45.7)

v = 29.9 m/s

b) Drawing a free body diagram, there are three forces on the snowboarder. Normal force up, gravity down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N - mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Therefore, the snowboarder's final speed is:

v² = v₀² + 2a(x - x₀)

v² = (29.9)² + 2(-9.8×.102) (10 - 0)

v = 29.6 m/s

Using energy instead:

KE = KE + W

1/2 mv² = 1/2 mv² + F d

1/2 mv² = 1/2 mv² + mgμ d

1/2 v² = 1/2 v² + gμ d

1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)

v = 29.6 m/s

c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.

∑F = ma

-mg sin θ = ma

a = -g sin θ

Therefore, the final height reached is:

v² = v₀² + 2a(x - x₀)

(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)

h = 44.7 m

Using energy:

KE = PE

1/2 mv² = mgh

h = v² / (2g)

h = (29.6)² / (2×9.8)

h = 44.7 m

3 0

4 years ago

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

what is the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board that is sand

Len [333]

F = normal force by each board on each side

W = weight of the board in between acting in down direction = 95.5 N

f = frictional force in upward direction by each board

$\mu$ = coefficient of friction = 0.663

Using equilibrium of force in Upward direction

f + f = W

f = W/2

f = 95.5/2 = 47.75 N

frictional force is given as

f = $\mu$ F

47.75 = (0.663) F

F = 72.02 N

4 0

3 years ago

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

Lunna [17]

Answer:

statement B is true

Explanation:

since same force is applied by the compressed spring on both masses so their rate of change of momenta must be same and since the lighter block has lesser mass so it must have greater velocity to have an equal change in momentum as of heavier mass.

By relation: $KE=\frac{(mv)^{2} }{2m}$

$KE lighter=\frac{(mv)^{2} }{2(1)}$ , $KEheavier=\frac{(mv)^{2} }{2(2)}$

comparing momenta of above two equations we get

KElighter (2) = KEheavier (4)

KElighter = 2 KEheavier

7 0

4 years ago