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makvit [3.9K]
3 years ago
13

How do carbon-12 and carbon-13 differ?

Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

They have different amounts of neutrons.  They have different mass numbers.

Leokris [45]3 years ago
4 0

They have different amount of neutrons

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Explain why the total positive charge in every atom of an element is always the<br> same.
geniusboy [140]

Answer:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral.

Explanation:

3 0
2 years ago
Read 2 more answers
How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?
svet-max [94.6K]
We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!
8 0
3 years ago
An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m
Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement  0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

     = \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0
3 years ago
It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon
trapecia [35]

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

3 0
3 years ago
A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the
kompoz [17]

' A ' and ' D ' are both correct statements.

3 0
2 years ago
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