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2 years ago

Plz help, and show work

Physics

1 answer:

pentagon [3]2 years ago

8 0

Answer:

1. 2.67 s

2. 0.1 m/s²

Explanation:

1. Determination of the time taken for the penguin to fall.

Height (h) of cliff = 35 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

35 = ½ × 9.8 × t²

35 = 4.9 × t²

Divide both side by 4.9

t² = 35 / 4.9

Take the square root of both side

t = √(35 / 4.9)

t = 2.67 s

Thus, it will take 2.67 s for the penguin to fall onto the head of a napping polar bear.

2. Determination of the acceleration of the penguin.

Initial velocity (u) = 0 m/s.

Final velocity (v) = 2 m/s.

Time (t) = 20 s

Acceleration (a) =?

a = (v – u)/t

a = (2 – 0)/ 20

a = 2 / 20

a = 0.1 m/s²

Thus, the acceleration of the penguin is 0.1 m/s²

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Svetllana [295]

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2 years ago

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umka21 [38]

Answer:

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Explanation:

5 0

2 years ago

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

5 0

2 years ago

PLEASE HELP

GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

2 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

2 years ago