Ask question

Ask question

All categories

4 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]4 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]4 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

ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude

Setler [38]

Answer:

a) $1.2\times 10^2\ N$

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2$

The acceleration of the bicycle and rider is -1.5 m/s²

Force

$F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N$

The magnitude of the average force needed to bring the bicycle and its rider to a stop is $1.2\times 10^2\ N$

3 0

3 years ago

Am 1 years old UwU (JK 15)

igomit [66]

Answer:

so am i ._. 15 not 1 ofc lol

Explanation:

7 0

3 years ago

Can someone please help with this problem?

andriy [413]

no question please so what's the problem

4 0

3 years ago

(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat

vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have

∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore

ñ = ⅟4(2·i +3·j -√3·k)

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0

3 years ago

2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral

AnnyKZ [126]

Answer:

b. negative

Explanation:

neutrons have a negative charge and protons have a proton has a positive charge

3 0

3 years ago

Read 2 more answers