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viva [34]
3 years ago
14

If Sarah melts 10g of tin. What mass of melted tin would she have at the end of the experiment?​

Physics
2 answers:
Nastasia [14]3 years ago
7 0

Answer:

10g

Explanation:

As the Law of Conservation of Mass states that " Mass can neither be created nor be destroyed in a chemical reaction".

Though melting of tin isn't a chemical change, the same logic is applied here...

Hence,

The mass of tin will be 10 g itself...

astra-53 [7]3 years ago
5 0

Answer:

10g

Explanation:

The experiment is about melting of tin. When solid tin is melted, it retains the amount of substance it originally has and we expect it to still be 10g.

Mass is the quantity or amount of matter contained in a substance. For most chemical processes, the law of conservation of matter is always succinctly observed.

The law states that "matter is neither created nor destroyed in a chemical reaction or process". Although melting is a phase change, we can adapt this law to the process. Owning to no loss of matter in the melting procedure, the amount of substance remains the same.

Therefore, we are left with about the same mass of substance we started with which is 10g.

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What type of magnetism is present in a magnet that you use to place a photograph on the refrigerator?
Vika [28.1K]

Answer:

C) Contact

Explanation:

The magnet requires almost direct <u>contact</u> with the fridge to start its magnetic properties.

7 0
3 years ago
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A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0
3 years ago
The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
1) how does reflection differ from diffraction?
Evgesh-ka [11]

Answer: (1) The correct answer is A.

(2) The correct answer is D.

Explanation:

(1)

Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.

Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.

Therefore, the correct answer is A.

(2)

Amplitude is the maximum displacement in the medium from the rest position.

The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.

Therefore, the correct option is D.

5 0
3 years ago
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Circuit has a single pathway for the electrons of flow.<br> series<br> parallel
Bad White [126]
Series circuit!!!!!!!!!!
5 0
3 years ago
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