What is the ability of the body to function successfully and efficiently during

an object is moving forward with a constant velocity. which statement about this object must be true?

love history [14]

Answer:

• Its acceleration is zero

Explanation:

from definition of acceleration:

${ \rm{acceleration = \frac{change \: in \: velocity}{time} }} \\ \\ { \rm{acceleration = \frac{ \delta \: (velocity)}{time} }}$

• since velocity is constant, change in velocity is zero

${ \boxed{ \rm{ \delta \: (velocity) = 0}}}$

• therefore:

${ \rm{acceleration = \frac{0}{time} }} \\ \\ { \boxed{ \tt{acceleration = 0 \: m {s}^{ - 2} }}}$

4 0

3 years ago

Rank in order the following according to their speeds, from fastest to slowest: (i) 425-nm-wavelength light through a pane of gl

kotegsom [21]

Answer:

v > ii > iii > i > iv

Explanation:

The speed of light does not depend on the wavelength of light rather the medium through which it passes through.

v) The speed of light in a vacuum is maximum = 300000 km/s

ii) The speed of light in air = 2999910.026 km/s

iii) The speed of light in water = 2255639.097 km/s

i) The speed of light in glass = 1973684.210 km/s

iv) The speed of light in diamonds = 1239669.421 km/s

Hence the ranking will be

v > ii > iii > i > iv

5 0

3 years ago

Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor

TEA [102]

Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

6 0

4 years ago

If the western component of the wind is half of the south, what is the angle of the wind with the south?

den301095 [7]

Answer:

This above a triangle that models our situation.

Explanation:

We have a two componens., since we have a western componet and southern component. One travel in a southern direction. and the other travel in the west.

Let the component that travel in the south be the length of a.

According to the problem, the westard component is half of that so let that length be a/2.

Now we must find the angle of the wind in the South.

This means that what is angle that is opposite of the western componet because that angle is the most southward angle. So know we apply the tan property.

$\tan(x) = \frac{opp}{adj}$

Our side opposite of the angle we trying to find is the western component and the side adjacent to it is the southern component. Also remeber since western and Southern negative displacements, we have

$\tan(x) = \frac{ - \frac{a}{2} }{ - a}$

$\tan(x) = - \frac{a}{2} \times - \frac{1}{a} = \frac{1}{2}$

Now we take the arctan or inverse tan of 1/2.

$\tan {}^{ - 1} ( \frac{1}{2} ) = 26.57$

3 0

2 years ago

Read 2 more answers

During aerobic activity, if your heart rate is lower than the lower limit, you are ___________________.

soldier1979 [14.2K]

Not pushing yourself hard enough is the answer since your heart rate doesn't even hit your lower minimum.
Pushing yourself to the limit is at your max heart rate.
Just at the right spot is at your max heart rate.
Pushing yourself too hard is above your max heart rate.

4 0

4 years ago