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3 years ago

12

What happens to temperature as a substance melts and heat energy is used to break the connections between molecules, until all m

elting is complete? increases steadily decreases steadily stays the same

1 answer:

zaharov [31]3 years ago

3 0

Answer:

Stays the same.

Explanation:

The temperature does not change just because something melts.

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A device for exercising the upper leg muscle is shown above, together with a schematic representation of an equivalent lever sys

yuradex [85]

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4 0

3 years ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

What is the empirical formula of metaldehyde, C8H16O4?

inessss [21]

Answer:

C8H16O4

Explanation:

ok??

7 0

2 years ago

A fluid flows along the x axis with a velocity given by V = ( x / t ) ˆ i , where x is in feet and t in seconds. (a) Plot the sp

umka21 [38]

Answer:

c)

V_local = -x/t^2

V_convec = x/t^2

d)

a = V_local + V_convec = 0

e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases

Explanation:

Hi!

You can see plots for a) and b) attached on this document

c)

The local acceleration is just teh aprtial derivative of the velocity with respect to t:

$\frac{dV}{dt} = \frac{d}{dt} \frac{x}{t}=- \frac{x}{t^2}$

And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:

$\vec{v} \cdot \nabla \vec{v} = v ( \frac{dv}{dx} ) =\frac{x}{t} \frac{1}{t} = \frac{x}{t^2}$

d)

Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign

e)

This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases, the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero

3 0

3 years ago

Please help with #8... Thanks in advance

schepotkina [342]

If the acceleration is constant, and the starting velocity is zero, the relationship between the acceleration of a falling body (a), the time it takes to fall (t), and instantaneous velocity when it hits the ground (v) is:

the general equation of acceleration is:

vf = vi + at

assuming the initial velocity (vi) is zero, the equation becomes:

vf = at

v = at

6 0

3 years ago