Which situation is an example of transferring heat by means of convection?

A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show

Lady bird [3.3K]

Answer:

change in height is 1.664 mm

Explanation:

Given data

drops = 3.00 mm

diameter = 5.00 cm = 0.05 mm

decrease = 350 cm^3

temperature = 95°C to 44.0°C

to find out

the decrease in millimeters in level

solution

we will calculate here change in volume so we can find how much level is decrease

change in volume = β v change in temp ...............1

here change in volume = area× height

so = $\pi$ /4 × d² h

so we can say change in volume = $\pi$ /4 × d² × change in height .......2

so from equation 1 and 2 we calculate change in height

( β(w) -β(g) )× v× change in temp = $\pi$ /4 × d² × change in height

change in height = 4 × ( β(w) -β(g) ) v× change in temp / $\pi$ /4 × d²

put all value here

change in height = 4 × ( 210 - 27 )(350 ) $10^{-6}$ × (95-44) / $\pi$ /4 × 0.05²

change in height is 1.664 mm

5 0

3 years ago

Which of the following statement(s) is

alexgriva [62]

Answer:

Explanation:

A rectifier is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. Rectifiers have many uses including as components of power supplies and as amplitude modulation detectors (envelope detectors) of radio signals. Rectifiers are most commonly made using solid state diodes but other type of components can be used when very high voltages or currents are involved. When only a single diode is used to rectify AC (by blocking the negative or positive portion of the waveform), the difference between the term diode and the term rectifier is simply one of usage. The term rectifier describes a diode that is being used to convert AC to DC. Most rectifier circuits contain a number of diodes in a specific arrangement to more efficiently convert AC power to DC power than is possible with only a single diode.

5 0

3 years ago

Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet

Svetllana [295]

Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s

6 0

3 years ago

which characteristic of northern renaissance artworks do albrecht dürer and pieter bruegel the elder share? a. they rarely inclu

VladimirAG [237]

According to my assumptions,the answer closest to the perfection is :

c. they show royalty instead of peasants.

7 0

4 years ago

a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra

Nikolay [14]

The dimensions of the rectangle are:

l = 50 m

b = $100/\pi$ m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is = $\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l$