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3 years ago

8

Which nutrient is the first (easiest) to be broken down and absorbed for energy use in the body? Water Carbohydrates Fats Protei

ns
Whole grains are considered a good source of carbohydrates.
a. True
b. False

1 answer:

Serjik [45]3 years ago

5 0

Whole grains are good carbs so it would be true.
And the easiest would be carbohydrates.

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In each case, lifting or pushing, why must you exert a force to move the object? Q1-2: How much more effort is required to lift

musickatia [10]

Answer:

Newton's first law applies. An object at rest will stay that way until a force is applied.
Any amount of effort can be applied to any amount of mass (in the ideal case). The question is not sufficiently specific.

Explanation:

A force is required to move an object because the object will stay at rest until a force is applied.

__

The effort required to lift or push two masses instead of one depends on the desired effect. For the same kinetic energy, no more effort is required. For the same momentum, half the effort is required for two masses. For the same velocity, double the effort is required.

4 0

3 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

3 years ago

A 4.0kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially

Kisachek [45]

a)

We use the formula :

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values in:

4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f

Calculating this we get:

32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f

Rearrange for v2f:

v2f = $\frac{32.0kg*m/s}{4.0kg}$

This gives us 8.0 m/s as the final velocity of the second ball.

b)

Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.

This means we use the formula:

Ek = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2} *m*v^{2}$ = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2}*m*v^{2}$

Substituting in values:

Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2

This simplifies to:

Ek= 128J + 0J = 0J + 128J

This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.

6 0

2 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

2 years ago

Reyna kicks a ball, which then rolls across a grassy field. Which is the best explanation for why the ball eventually comes to a

nataly862011 [7]

Short Answer- Friction

Explanation- There is friction between the grass and the ball. Friction is a type of force that limits motion between two surfaces that are touching. Friction is the force that one surface exerts on another with the two surfaces rubbing together.

This is the answer, hope it helps :)

8 0

3 years ago