Where is A what is the answer?

In which of these samples do the molecules most likely have the most kinetic energy? (2 points)

Vinvika [58]

D, water vapor. Gaseous state would have more kinetic energy, they are moving faster. If you have to compare the same state, then higher temperature would have the higher kinetic energy. But if you have solid and liquid at the same temperature - then liquid would have more.

4 0

3 years ago

A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that

pav-90 [236]

Answer:

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y $_{b}$ ( V₁² - V₂²)) / ( 2g( ηbμ + $f_{rl}$ ± sin∅ $_{g}$ ))

now given that the tracked is levelled, ∅ $_{g}$ = 0, also Y $_{b}$ = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592) = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore, the unloaded braking efficiency is 84.6 %

7 0

3 years ago

A ball is thrown vertically upward with a speed of 1.86

Inessa05 [86]

Answer:

t = 1.09 s.

Explanation:

This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.

$y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0$

This quadratic equation can be solved using determinant.

$\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s$

Of course, we will choose the positive time.

4 0

3 years ago

a car takes off from rest and covers a distance of 80m on a straight road in 10s.calculate the magnitude of its acceleration

olasank [31]

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Here's the solution :

Let's find the final velocity :

$\dfrac{displacement}{time}$

$\dfrac{80}{10}$

$8 \: \: ms {}^{ - 1}$

Initial velocity (u) = 0 (cuz it started from rest)

Final velocity (v) = 8 m/s

Time taken (t) = 10 sec

now, we know that :

$acceleration = \dfrac{v - u}{t}$

$\dfrac{8 - 0}{10}$

$\dfrac{8}{10}$

$0.8 \: \: m {s}^{ - 2}$

Acceleration = 0.8 m/s²

7 0

3 years ago

A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction

Nezavi [6.7K]

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

$F_{n}$ = Normal force on the block by the ramp

$f_{k}$ = kinetic frictional force

Force equation perpendicular to ramp surface is given as

$F_{n} = mg Cos65$

Kinetic frictional force is given as

$f_{k} = \mu F_{n}$

$f_{k} = \mu mg Cos65$

Force equation parallel to ramp surface is given as

$mg Sin65 - f_{k} = ma$

$mg Sin65 - \mu mg Cos65 = ma$

$g Sin65 - \mu g Cos65 = a$

$(9.8) Sin65 - (0.17) (9.8) Cos65 = a$

$a = 8.2$ m/s²

4 0

4 years ago