Ask question

Ask question

All categories

3 years ago

11

The study of anatomy and physiology assumes and describes a healthy body. Select the description below that does NOT explain why

this approach is useful.

1 answer:

galben [10]3 years ago

5 0

Answer:

Explanation:

The question is incomplete. Here is the complete question.

The study of anatomy and physiology assumes and describes a healthy body. Select the description below that does not explain why this approach is useful.

a) a healthy body provides a common standard to compare to.

b) Study of a healthy body is less intimidating and more familiar to new students.

c) A healthy body establishes what normal is.

d) Study of a healthy body provides a foundation for a more complete understanding of all human bodies.

b) option is correct. That study of a healthy body is less intimidating and more familier to new students.

You might be interested in

An object of 4 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules.

11Alexandr11 [23.1K]

Answer:

The answer to your question is Ke = 72 J

Explanation:

Kinetic energy depends on the speed of and object and its mass.

Data

mass = m = 4 kg

speed = v = 6 m/s

distance = d = 8 m

Kinetic energy = ke = ?

Formula

Ke = (1/2) mv²

Substitution

Ke = (1/2) (4)(6)²

Simplification

Ke = (1/2)(4)(36)

Ke = (1/2)(144)

Ke = 72 Joules

Result

Ke = 72 J

3 0

2 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

Paladinen [302]

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

$F_b = \rho * V * g$

$F_b = 1000*0.10 * 0.08 * h * 9.8$

now by above force balance equation we can write

$m*g = F_b$

$0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8$

$0.040 = 8h$

$h = 5 * 10^{-3} m$

so boat will sunk by total 5 mm distance

8 0

3 years ago

How are a concave lens and a convex lens alike? How are they different?

Verdich [7]

ANSWER

A convex lens acts a lot like a concave mirror. ... A concave lens acts a lot like a convex mirror. Both diverge parallel rays away from a focal point, have negative focal lengths, and form only virtual, smaller images.

5 0

3 years ago

The small ball of mass m and its supporting wire become a simple pendulum when the horizontal cord is severed. Determine the rat

natali 33 [55]

Answer:

See the attached image and the explanation below

Explanation:

We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.

These free body diagrams can be seen in the attached image.

First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).

In this way we get the T-wire tension equation, before cutting.

Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).

It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.

Finally with equations 1 and 2 we can find the K ratio.

5 0

3 years ago