Ask question

Ask question

All categories

2 years ago

7

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

inner cross-sectional area of 0.012 mm2. the coefficient of volume expansion of mercury is 1.8 × 10-4 k-1. if the expansion of the glass is negligible, how much will the mercury rise in the capillary tube when the temperature rises from 5°c to 35°c if the bulb was full at 5°c?

2 answers:

Nadya [2.5K]2 years ago

7 0

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm

Answer: 4.5 cm

liraira [26]2 years ago

6 0

The mercury will rise 4.5 cm in the capillary tube.

$\texttt{ }$

<h3>Further explanation</h3>

This problem is about Density.

Density is the ratio of mass to the volume of the object.

$\large {\boxed {\rho = \frac{ m }{ V } } }$

<em>ρ = density of object ( kg / m³ )</em>

<em>m = mass of object ( kg )</em>

<em>V = volume of object ( m³ )</em>

$\texttt{ }$

The volume expansion of object could be calculated using this following formula.

$\boxed {V = Vo ( 1 + 3 \alpha \Delta t )}$

where:

<em>V = final volume after expansion ( m³ )</em>

<em>Vo = initial volume before expansion ( m³ )</em>

<em>α = coefficient of expansion ( /K )</em>

<em>Δt = change in temperature ( K )</em>

<em>Let's tackle the problem!</em>

$\texttt{ }$

<u>Given:</u>

initial volume = Vo = 0.100 cm³

cross-sectional area = A = 0.012 mm² = 0.012 × 10⁻² cm²

coefficient of volume expansion of mercury = γ = 1.8 × 10⁻⁴ /K

change in temperature = Δt = 35 - 5 = 30°C

<u>Asked:</u>

the rise of mercury column = h = ?

<u>Solution:</u>

$h = \Delta V \div A$

$h = (V_o \gamma \Delta t) \div A$

$h = (0.100 \times 1.8 \times 10^{-4} \times 30) \div (0.012 \times 10^{-2})$

$h = 4.5 \texttt{ cm}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Expansion

You might be interested in

A 20 watt lightbulb uses 20 Joules of energy every second.A person expends 50 watts of energy per stair when climbing up stairs.

Pachacha [2.7K]

20W = 20 J/s

Energy expended during climbing stairs = 50 W of energy/stair = 50J/stair

For 20 stairs, Total energy = 50x20 = 1000 J

This can light bulbs for, T= 1000J/20 J/s =50 seconds

6 0

2 years ago

A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f

snow_tiger [21]

Answer:

$2.5 \times 10^{11} N-m$ upwards

Explanation:

Torque is the vector cross product of the force and radial distance.

$\tau = rF sin \theta$

$\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m$

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0

2 years ago

An airplane flying at constant air speed from Indianapolis to St. Louis (assume they are directly east-west of each other) in ca

irga5000 [103]

Answer:

Explanation:

Suppose the distance between the two cities is D and the velocity in calm weather is V . The total time taken in two way travel is given by

Total distance / velocity

= 2 D / V

Suppose velocity of wind is v . Then in one way the velocity of airplane will become V + v and in the return trip its velocity will be V - v

Total time taken

= $\frac{D}{V+v} +\frac{D}{V-v}$

= $\frac{2DV}{V^2-v^2}$

= $\frac{2V^2D}{V(V^2-v^2)}$

= $\frac{2D}{V(1 - \frac{v^2}{V^2}) }$

= The denominator contains a factor

$1-\frac{v^2}{V^2}$

which is less than one so time calculated will be more than

2D / V

Hence in the second case time taken will be more .

7 0

2 years ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

2 years ago