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trapecia [35]
3 years ago
10

Why it is not easy to whril a hammer by a longer chain?

Physics
1 answer:
-Dominant- [34]3 years ago
6 0
<span>because with a longer chain you are giving the hammer momentum so it will take more strength to move</span>
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Ronch [10]
For your reference.For your reference.

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Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.
kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, 4.67\times 10^7

(b) The number of radon nuclei have decayed by this time will be, 3.6\times 10^8

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{3.82\text{ days}}

k=1.81\times 10^{-1}\text{ days}^{-1}

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.81\times 10^{-1}\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initially number of radon atoms  = 4.1\times 10^8

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}

a-x=4.67\times 10^7

Thus, the number of radon atoms will remain after 12 days is, 4.67\times 10^7

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = (4.1\times 10^8)-(4.67\times 10^7)

The number of radon nuclei have decayed = 3.6\times 10^8

Thus, the number of radon nuclei have decayed by this time will be, 3.6\times 10^8

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3 years ago
What is the difference between an atom in the ground state and an atom in an excited state
SIZIF [17.4K]
The atom in an excited state has more energy and is less stable than the atom in the ground state.
7 0
4 years ago
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1 point
marta [7]

Answer:

Let, R be the resistance of the heater wire. Since, two heater wires are of equal length their resistance is also same.

Hence, for series combination of resistances,

R

s

=2R

And for parallel combination of resistances,

R

p

=

2

R

Now, heat produced when they first connected in series is

H

s

=

R

s

V

2

where, V is voltage supplied to the heater.

H

s

=

2R

V

2

.....................................(1)

heat produced when they first connected in series is

H

p

=

R

p

V

2

H

p

=

2

R

V

2

H

p

=

R

2V

2

............................................(2)

From (1) and (2), we get

H

p

H

s

=

R

2V

2

2R

V

2

⇒

H

p

H

s

=

4

1

7 0
3 years ago
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