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2 years ago

Which of these experiments would make use of qualitative data?

Physics

2 answers:

Umnica [9.8K]2 years ago

6 0

A. A study of different surfaces to compare luster or shine.

xxTIMURxx [149]2 years ago

6 0

Answer: A. a study of different surfaces to compare luster or shine

Explanation:

A qualitative data can be define as the data that can only be observed and recorded but it cannot be counted in numbers. This type of data cannot be measured in units. For example, color.

A is the correct option. This is because of the fact that luster and shine can only be observed it cannot be measured in units and counted in numbers. It just represents the physical observable property of a material.

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B. The silica cylinder of a radiant wall heater is 0.6 m long

SIZIF [17.4K]

So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

ε = emissivity = 1 (since we are not given),
σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
A = surface area of cylindrical wall heater = 2πrh where
r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
h = length of heater = 0.6 m, and
T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

6 0

2 years ago

An airplane is flying in the direction 10° east of south at 701 km/hr. Find the component form ofthe velocity of the airplane, a

solniwko [45]

Answer:

The component form will be;

In the x-axis = 121.73 due west

In the y-axis = 690.35 due south

Explanation:

An image of the calculation has been attached

7 0

2 years ago

WILL GIVE YOU BRAINLIST IF YOU ANSWER Which of the following characteristics of the Arctic rabbit is specifically an adaptation

andrew-mc [135]

Answer:

Hold active layer of soil in place; act as producers in ecosystem

5 0

3 years ago

The unit of current, the ampere, is defined in terms of the force between currents. Two 1.0-meter-long sections of very long wir

Paha777 [63]

Answer:

1.6 x $10^{-7}$ N

Explanation:

If the force between two actual wires has this value, the current is defined to be exactly 1 A.

The force between two parallel wires carrying current can be defined as,

F= μ $I_{1}$ $I_{2}$ L/ 2πd

where,

current $I_{1}$ = 1Amp

current $I_{2}$ = 2amp

Length 'L'= 1m

distance 'd'= 2.5m

permeability of free space 'μ'= 4πx $10^{-7}$ N/m

Putting the above values in the equation,

F=( 4πx $10^{-7}$ x 1 x 2 x 1 )/ 2πx2.5

F= 1.6 x $10^{-7}$ N

8 0

3 years ago

Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges

poizon [28]

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0

2 years ago