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iVinArrow [24]
3 years ago
9

If a rock is thrown 0.8 meters into the air, How fast was it thrown

Chemistry
2 answers:
PIT_PIT [208]3 years ago
5 0

Answer:

Speed, s = 3.95 m/s

Explanation:

It is given that, a rock is thrown 0.8 meters into the air. We have to find the velocity with which it is thrown.  

Using conservation of mechanical energy as :

K_i+P_i=K_f+P_f

Initial potential energy at the ground is 0 and at maximum height the final kinetic energy is 0.

So,

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

putting g = 9.8 m/s² and h = 0.8 m

v=\sqrt{2\times 9.8\times 0.8}    

v = 3.95 m/s

Hence, it was thrown with a speed of 3.95 m/s

sergiy2304 [10]3 years ago
3 0
V=0 m/s, u=?, a = 9.8 m/s² and s = 0.8 m
u²=2×9.8×0.8=15.68
<span>u=3.959 ≈ 3.96 m/s

</span>
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Answer:Static electricity works because objects which are otherwise "neutral" (in other words, objects with no net charge) can be polarized. An electric field, like one caused by a nearby charged object, can cause the charges inside of a neutral object — the protons and electrons — to move around a tiny bit.

Explanation:

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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5
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Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}

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  = 1.21\times 10^{-4}years^{-1}

t = age of sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = \frac{70}{100}\times 100=70

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}

t=2948years

Thus the fossil is 2948 years old.

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Function of starch solution​
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What is the specific name for the rock formed when magma cools
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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