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kirill [66]
3 years ago
8

What is the amount of matter in an object is called?​

Physics
2 answers:
HACTEHA [7]3 years ago
6 0

Answer: Mass is the amount of mass an object contains. Mass is measured using a scale. Volume is the amount of space matter takes up.

hope this help!! :) :)

Sedbober [7]3 years ago
3 0

Answer:

mass

Explanation:

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It extracts energy from food for the cell.
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Which of the following is considered an energy source and a building block for the human body?
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032. A 25.0 kg bicycle is traveling at a speed of 10.0 m/s.
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A: 4
B: 7
C. 3
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7 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Leya [2.2K]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

6 0
3 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
4 years ago
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