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Zepler [3.9K]
3 years ago
5

A class of students performed the heat transfer experiment shown in the picture. In the experiment, three plastic dishes were pa

rtially filled with equal amounts of dark soil, light soil, and water. Thermometers were placed in each dish, and the dishes were placed under a heating lamp. The substances were left under the heating lamp for ten minutes, and their temperatures were observed and recorded. After recording the temperatures of the substances as they were heating, the students turned the lamp off and recorded the temperatures of the substances as they cooled for ten minutes.
Which method of heat transfer is illustrated in this experiment? Explain your answer

Physics
1 answer:
balandron [24]3 years ago
6 0

The method of heat transfer used in the experiment is radiation.

Radiation is a method of heat transfer in which there is no material medium that conveys heat. Heat travels through empty space and causes the temperature of objects to rise.

In this case, there was no material medium that conveyed heat from the lamp to the dishes containing the plants. Heat from the lamp traveled through space and reached the dishes. This is an example of heat transfer via radiation.

Learn more: brainly.com/question/12494464

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Using an elastic cord, the astronaut in the black shirt pulls the astronaut with the red shirt back towards him before releasing
pishuonlain [190]

Answer:

50 T-shirt Design Ideas That Won't Wear Out - 99Designshttps://99designs.com › blog › creative-inspiration › t-s...

Whatever message your t-shirt needs to send, we've got 50 t-shirt design ... Illustrated chicken astronaut t-shirt ... of the challenges associated with a t-shirt for a business is getting people to wear it. ... shirt is a great way to create memorabilia that participants can look back on ... Graphic tees are what shirts were made for.

Explanation:

8 0
3 years ago
A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from
ankoles [38]

Answer

current density is given by

I = \int J.da

where J = σ E

I = \int J.da\\ =\int \sigma E .da\\ = \sigma E \int da \\ =\sigma E \int_0^{2\pi} rL

now,

E=\dfrac{I}{2\pi r L \sigma}

a) Electric field strength at the inner surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi a L \sigma}

E=\dfrac{27}{2\pi\times 0.005\times 0.1\times 10^7}

      E = 8.59 x 10⁻⁴ V/m

b) Electric field strength at the outer surface of an iron cylinder

a = 0.5 cm = 0.005 m             b = 2.3 = 0.023

L = 10 cm = 0.1 m                     I = 27 A

E=\dfrac{I}{2\pi b L \sigma}

E=\dfrac{27}{2\pi\times 0.023\times 0.1\times 10^7}

      E =1.87 x 10⁻⁴ V/m

7 0
3 years ago
If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will Group of answer c
Sindrei [870]

Answer:

The correct answer is Option A (decrease).

Explanation:

  • According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.
  • Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.

Other given choices are not connected to the given query. Thus the above is the right answer.

5 0
3 years ago
Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th
Sergio039 [100]

Answer:

The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell q_{inn}=3.50\times10^{-8}\ C

Charge on outer shell q_{out}=1.60\times10^{-8}\ C

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Where, q = charge

k = constant

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}

E=2.38\times10^{4}\ N/C

The total charge enclosed  by a radial distance 20.5 cm

The total charge is

q=q_{inn}+q_{out}

Put the value into the formula

q=3.50\times10^{-8}+1.60\times10^{-8}

q=5.1\times10^{-8}\ C

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}

E=1.09\times10^{4}\ N/C

Hence, The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

7 0
3 years ago
A solution of phosphoric acid was made by dissolving 10.0 g H3PO4 in 100.0 mL water. The resulting volume was 104 mL. Calculate
OLEGan [10]

Answer:

density is 1.057 g/mL

mole fraction is 0.0180

molarity is 0.98 mol/L

mole fraction is 0.0180

Explanation:

Given data

acid mass = 10 g

water volume = 100 mL

total volume = 104 mL

density = 1 g/cm³

to find out

the density, mole fraction, molarity, and molality

solution

first we calculate the density that is = total mass g / volume of solution mL

total mass = mass of H3PO4 + water  mass

so water mass = density x volume

water mass = 100 ml x 1.0  g/cm3

water mass = 100 g  

so total mass  = 110.00 g

so that

density = total mass g / volume of solution mL

density = 110 / 104 = 1.057 g/mL

now we calculate no of moles in solvent i.e = mass  H2O / mlar mass H2O

no of moles in solvent = 100/18.015 = 5.55 moles

and no of moles in solute i.e = mass of H3PO4 / mlar mass in H3PO4

moles in solute i.e = 10/ 97.994 = 0.102 moles

so total moles is  5.55  + 0.102 = 5.652 moes

so now mole fraction = no of moles in solute / total moes

mole fraction = 5.55 / 5.65

mole fraction is 0.0180

now we calculate first

mole fraction in solute and solvent that is

mole fraction in solute = no of moles in solute/ total moles

mole fraction in solute = 0.102 /5.65

mole fraction in solute is 0.0180

and mole fraction in solvent that = no of moles in solvent/ total moles

mole fraction in solvent that = 5.55/ 5.65

mole fraction in solvent that is 0.982

so molarity = no of moles of solute / volume

molarity = 0.102/0.104

molarity is 0.98 mol/L

and molality  is = no of moles of solute / mass

molality = 0.102 / 100

molality  is  1.02 mol/kg

6 0
3 years ago
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