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BARSIC [14]
3 years ago
15

The function h(t) = 162 +64t models the height, in feet, of a ball t seconds after it was kicked into the air. Which statement i

s true?
A. The ball reaches its maximum height 4 seconds after it was kicked.
O B. The ball reaches its maximum height 8 seconds after it was kicked.
O C. The ball takes 2 seconds to hit the ground.
D. The ball takes 4 seconds to hit the ground.
Physics
1 answer:
melamori03 [73]3 years ago
3 0

Answer:

the ball didn't not reach the Maximum height because of the time interval

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The visible spectrum refers to the portion of the electromagnetic spectrum that we ________.
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<h2>Answer: can see</h2>

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately.  Which  means this part of the spectrum is located between ultraviolet light and infrared light.  

Note the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

Therefore:

<h2>The visible spectrum refers to the portion of the electromagnetic spectrum that <u>we </u><u>can see</u></h2>
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The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will
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Greater than 25 . 2.5 x 10 = 30
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3 years ago
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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
kakasveta [241]

Answer:

W=4037.36\ J

Explanation:

Given:

mass of ice melted, m=3.3\times10^{-2}\ kg

time taken by the ice to melt, t=5\ min=300\ s

latent heat of the ice, L=3.34\times 10^5\ J

Now the heat rejected by the Carnot engine:

Q_R=m.L

Q_R=0.033\times 3.34\times 10^5

Q_R=11022\ J

Since we have boiling water as hot reservoir so:

T_H=373\ K

The cold reservoir is ice, so:

T_L=273\ K

Now the efficiency:

\eta=1-\frac{T_L}{T_H}

\eta=1-\frac{273}{373}

\eta=26.81\%

Now form the law of energy conservation:

Heat supplied:

Q_S-W=Q_R

where:

Q_S= heat supplied to the engine

Q_S-\eta\times Q_S=Q_R

Q_S(1-\eta)=Q_R

Q_S=\frac{11022}{1-0.2681}

Q_S=15059.36\ J

Now the work done:

W=Q_S-Q_R

W=15059.36-11022

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8 0
3 years ago
Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's f
Lerok [7]

Answer:

A.a=203.14\ \frac{m}{s^2}

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

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a=203.14\ \frac{m}{s^2}

So the acceleration  while stopping will be a=203.14\ \frac{m}{s^2}.

Lets take distance travel before come top rest is s

We know that

v^2=u^2-2as

0=284.4^2-2\times 203.14\times s

s=397.6 m

So the distance travel while stopping is 397.6 m.

8 0
3 years ago
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Hurry please I don’t have long for this for a test !!!
Anna35 [415]
I think the Answer is b
7 0
3 years ago
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