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lesya692 [45]
3 years ago
14

An object is in free fall. After 10 seconds calculate the distance fallen? *

Physics
2 answers:
ch4aika [34]3 years ago
5 0

*For free fall,

Initial velocity (u) = 0

g = 9.8m/s²

time(t) = 10s

Using the second equation of free fall;

s = ut + 1/2gt²

where 's' is the distance fallen.

s = 0(10) + 1/2(9.8)(10)²

s = 1/2(9.8)(100)

s = 490m.

*Initial horizontal velocity (Ux) = 3m/s

Initial vertical velocity (Uy) = 0

since it did not start with any vertical velocity.

Using the second equation of vertical downward motion:

h = Uyt + 1/2gt²

where h is the height of the cliff = 60m.

g = 9.8m/s²

60 = 0(t) = 1/2 (9.8)(t²)

60 = 4.9t²

t = √(60/4.9) = 3.5s.

Citrus2011 [14]3 years ago
3 0

Answer:

9.81 × 10 = 98.1 meters

vertical displacement is s=1/2 at^2 + vt

initial vertical velocity is 0 so s=1/2 at^2

a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2

solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.

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Imagine a raindrop starting from rest in a cloud 2 km in the air. If it fell with no air friction at all, it would accelerate to
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Answer:

2) 433 mph

Explanation:

The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:

v^2 = u^2 + 2ad

where

v is the final velocity

u = 0 is the initial velocity (the raindrop starts from rest)

a = g = 9.8 m/s^2 is the acceleration due to gravity

d = 2 km = 2000 m is the distance covered

Solving for v,

v=\sqrt{u^2 +2gd}=\sqrt{0^2+2(9.8 m/s^2)(2000)}=198 m/s

And keeping in mind that

1 mile = 1609 metres

1 hour = 3600 s

The speed converted into miles per hour is

v=198 \frac{m}{s}\cdot \frac{3600 s/h}{1609 m/mi}=433 mph

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3 years ago
Explain why average velocity in one dimension can be positive or negative.
Softa [21]
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since velocity =  \frac{change in position}{time}, and time is always positive, velocity will be positive in one direction and negative in the other direction.
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3 years ago
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A honey bee flaps its wings 200 times per second. How much time is required for one wingbeat? Give your answer in milliseconds.
kakasveta [241]
A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.

Therefore, a bee flaps its wings over 200 times in about 1ms
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3 years ago
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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

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2 years ago
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