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3 years ago

In a collision, a car of mass 1000kg travelling at 24m/s comes to rest in 1.2s calculate

Physics

1 answer:

kompoz [17]3 years ago

8 0

Answer:

a. 24,000

b. -20,000 N

Explanation:

a. p = m•v

1,000•24 = 24,000

1,000•0 = 0

∆p = 24,000

b. F = m•a

a = ∆v/∆t

a = -24/1.2

a = -20 m/s²

F = 1,000•(-20)

F = -20,000

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An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

4 years ago

The first-order decomposition of cyclopropane has a rate constant of 6.7 × 10–4 s–1. If the initial concentration of cyclopropan

Ket [755]

Answer:

.864 M

Explanation:

For first order decomposition,

rate constant k = 1/t x ln a / (a - x )

given , a = 1.33 M , t = 644 s , k = 6.7 x 10⁻⁴ , a - x = ? = b( let )

6.7 x 10 ⁻⁴ = 1/644 x ln 1.33/b

ln 1.33/b = 6.7 x 10⁻⁴ x 644 = .4315

1.33 / b = e⁰ ⁴³¹⁵ = 1.5395

b = 1.33 / 1.5395 = .864 M.

7 0

4 years ago

What is the relationship between atoms and elements

pshichka [43]

Atoms and elements are pure substances.

5 0

4 years ago

Newton's laws of motion work well for ordinary situations on earth. However, these laws of motion do not work for all cases. Und

Alexandra [31]

Answer:

C: for objects at extremely fast speeds.

Explanation:

Newton's second law does not hold for extremely fast speeds, because then relativistic effects come into play, where Einstein's theory of special relativity is a more correct description.

The reason why F=ma does not hold for fast speeds is that, as an object moves faster and faster, the proportional relationship between force and acceleration does not hold. As an object moves faster and faster, it becomes harder and harder (requires more force) to accelerate it. because it gains mass as a virtue of its velocity (what's called relativistic mass).

For relativistic speeds, the correct modification of Newtons second law is:

$F=\frac{dp}{dt}$