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Nina [5.8K]
3 years ago
10

Albert dressed as TWO-FACE flips a coin up in the air at an upward velocity of 5.00 m/s. He fails to catch it on the way down an

d it falls down an 8.5 m deep well. Calculate the maximum height and maximum magnitude (could be positive or negative) of velocity of the coin.
Physics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

1.28 m, 14 m/s

Explanation:

At the maximum height, the velocity is 0.

Given:

a = -9.8 m/s²

v₀ = 5.00 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (5.00 m/s)² + 2(-9.8 m/s²) (x - 0 m)

x = 1.28 m

The maximum speed is at the bottom of the well.

Given:

a = -9.8 m/s²

v₀ = 5.00 m/s

x₀ = 0 m

x = -8.5 m

Find:

v

v² = v₀² + 2a(x - x₀)

v² = (5.00 m/s)² + 2(-9.8 m/s²) (-8.5 m - 0 m)

v = -13.8 m/s

Rounded to 2 sig-figs, the maximum speed is 14 m/s.

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What do nuclear fission and nuclear fusion have in common?
adoni [48]

Answer:

Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy

Explanation:

7 0
3 years ago
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cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat
Llana [10]

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

   r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

S = 2\pi \dfrac{27}{r} + \pi r^2

\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)

\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r

\frac{\mathrm{d} s}{\mathrm{d} r}=0

for least amount of material requirement.

\dfrac{54\pi }{r^2} = 2\pi r\\r=3m

hence radius comes out to be 3 m

for height put the value in the equation 1

so, height of the cylinder comes out to be 3m

3 0
3 years ago
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An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

3 0
3 years ago
I have to write an essay about Quantum Mechanics. I need to know everything about it. Please Help!
Studentka2010 [4]

ananı var ya ananı

kodumun abd lisi aptal cajhil özürlü

6 0
3 years ago
Read 2 more answers
Someone help me please.
Soloha48 [4]

1/2*2.8*x^{2}=2.8*9.81*1.50

X=5.425

Because K.E=1/2*M*V² and it is equal to P.E in the conservation of energy which means its K.E=P.E

P.E=m*g*h

So K.E equals 1/2*2.8*(5.425)²=41.20

8 0
3 years ago
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