Which model is the best example? (A, B, C, or D

A trunk rotation is a common dyamic flexibility assessment ture or false

True. It would be false if the statement was "trunk rotation is the most common static flexibility assessment."

So, you're answer should be "true". Hope that helped!

6 0

3 years ago

A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –

zhenek [66]

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

$W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J$

So, the work done by the given force is 18.88 J.

5 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

Why does the area around the equator stay about the same temperature year round?

Verdich [7]

Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.

3 0

3 years ago

What is the eulerian description of fluid motion how does it differ from the lagrangian description?

Alex_Xolod [135]

Kinematics : Study of motion

Fluid kinematics : study of how fluid flows and how to describe its motion.

There are two ways to describe fluid motion

one is Eulerian, where the variations are described at all fixed stations as a function of time.

the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.

DIFFRENCE BETWEEN LAGRANGIAN AND EULERIAN:

1.Both Lagrangian and Eulerian describes time variation.

2. Eulerian describes the rate of change in one point of space

Lagrangian descries rate of change of a property of material system.

To know more about the Lagrangian and Eulerian :\brainly.com/question/14944792

#SPJ4

3 0

1 year ago