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3 years ago

1. A drop of oil of volume 10-10 m spreads out on water to

Physics

1 answer:

ankoles [38]3 years ago

5 0

Answer:

318.3 nm

Explanation:

We approximate the circular film as a cylinder of height h and radius, r. Its volume V = πr²h. Since this volume equals the volume of the oil drop, the height of the circular film is thus h = V/πr²

V = 10⁻¹⁰ m³ and r = 10 m

Substituting into h, we have

h = 10⁻¹⁰ m³/π(10)²

= 0.3183 × 10⁻¹² m

= 3183 × 10⁻⁹ m

= 318.3 nm

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Layers of Earth's Atmosphere

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Answer:

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Troposphere

contains weather

contains life forms

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3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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2 years ago

A truck and a scooter moving with the same velocity colide. Why does the scooter fly off?

iragen [17]

We know, Momentum is the product of mass and velocity. As velocity is equal, momentum of Truck will be greater as mass of Truck is greater than the scooter. As momentum is greater from truck side, it will transfer the same amount to the scooter, so it flies away

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3 years ago