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3 years ago

8

Arun places a beaker of water on a hotplate and turns the hotplate on. The temperature in the room is 25 ºC. After the water hea

ts to 75 ºC, she adds pieces of ice to the beaker. Which mode of transfer of heat energy occurs in the water during this experiment? Justify your answer

1 answer:

Rzqust [24]3 years ago

5 0

Answer:

Thermal convection

Explanation:

When ice cubes are placed in water, ice cubes start melting. This is because water transfers its heat to ice cubes until the temperature becomes equal and the process is called "thermal Convection".

Thermal convection is the mode of transfer of heat energy that will occur when ice cubes will be added to the hot water and at the end of the experiment when both are at the same temperature, no ice cube will left.

Hence, the correct answer is "thermal convection".

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Convert 27,549 into scientific notation

dezoksy [38]

2.7549 x 10^4 is the answer I hope this helped u

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3 years ago

You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.

nika2105 [10]

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So $227=14\times a$

$a=16.214m/sec^2$

(b) Acceleration due to gravity $g=9.8m/sec^2$

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight $W=mg=14\times 9.8=137.2N$

So weight of the bag will be 137.2 N

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4 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

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3 years ago

-. A 2000 kilogram (kg) weather satellite is orbiting Earth. The satellite

melisa1 [442]

Answer:

H

Explanation:

I think it's H hope that helps

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2 years ago

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

3 years ago