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Natalija [7]
2 years ago
12

A flask is filled with 1.57 L (L: liter) of a liquid at 90.6 °C. When the liquid is cooled to 10.2 °C, its volume is only 1.38 L

. however. Neglect the contraction of the flask. What is the coefficient of volume expansion of the liquid?
Physics
1 answer:
nydimaria [60]2 years ago
8 0

The coefficient of volume expansion of the liquid = 1.97 × 10⁻³ /⁰C

<h3>What is the coefficient of volume expansion of the liquid?</h3>

The amount of volume that a substance expands by per unit of its original volume for each degree that its temperature rises is known as the coefficient of volume expansion.

As we know, The coefficient of volume expansion of the liquid

= change in volume/(original volume × temperature difference)

= (V₂ - V₁)/[V₁ × (T₂ - T₁)]

Change in volume = (V₂ - V₁)

Here, V₂ ( final volume) = 1.31 L

V₁  (initial volume)= 1.55 L

T₂ (final temperature) = 14.7 degrees

T₁ (initial temperature) = 96 degrees

The coefficient of volume expansion of the liquid

= (1.31 - 1.55) / [1.5 × (14.7- 96)]

= 1.97 × 10⁻³ /⁰C

Thus, the coefficient of volume expansion of the liquid = 1.97 × 10⁻³ /⁰C

To know more about coefficient of volume expansion refer to:

brainly.com/question/24042303

#SPJ1

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In this case, the object has no acceleration along horizontal direction, it has acceleration in vertical direction which is equal to the acceleration due to gravity of earth.

When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

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Question 14 (2 points)
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So, higher the frequency, smaller be the wavelength.

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3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
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Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

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3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

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