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pishuonlain [190]
3 years ago
7

You drop a rock off a bridge. The rock's height, h (in feet above the water), after t seconds is modeled by h = – 16 t2 + 541. W

hat is the height of the rock after 2 seconds?
Physics
1 answer:
marin [14]3 years ago
5 0
<span>h ( t) = h(1 sec) = -16t^2 + 541

so h (2 sec) = -16*(2)^2 + 541 = -64 + 541 = <span>477 ft

Therefore, </span></span>the height of the rock after 2 seconds is 477 feet.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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A 50 N girl climbs the flight of stairs in 3 seconds. How much work does she
kvv77 [185]
She uses 0 power because she simply has no gas left. She used all of it getting a 50 pc McDonald’s nugget with a side of ranch and a medium fry and a large water.
7 0
3 years ago
What is championship game of baseball called? The World Series The Super Bowl The World Cup​
emmasim [6.3K]

Answer:

The World Series

Explanation:

The Super Bowl is the championship American Football game, and the World Cup is the Soccer/Football game.

American Football and Football are different things. The first is what Americans call football, while the other is what Americans call soccer. It is confusing.

6 0
3 years ago
A ball is droped from a height of 16m how much time will pass before the ball hits the ground​
sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

g = 9.8m{s}^{-2}

By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

square root both sides, we obtain

\implies t= \sqrt{3.265}

t=1.8s

4 0
4 years ago
Please list the following for the Element, HYDROGEN
bagirrra123 [75]

Answer:

1 Proton, 1 Electron, No Neutrons

Group 1, Period 1

Gases

3 0
3 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
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