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ArbitrLikvidat [17]
3 years ago
5

A coffee filter of mass 1.4 grams dropped from a height of 4 m reaches the ground with a speed of 0.9 m/s. How much kinetic ener

gy Kair did the air molecules gain from the falling coffee filter? Start from the Energy Principle, and choose as the system the coffee filter, the Earth, and the air.
Physics
1 answer:
WITCHER [35]3 years ago
8 0

Answer:

The kinetic energy gained by the air molecules is 0.054313 J.                      

Explanation:

Given that,

Mass of a coffee filter, m = 1.4 g

Height from which it is dropped, h = 4 m

Speed at ground, v = 0.9 m/s

Initially, the coffee filter has potential energy. It is given by :

P=mgh\\\\P=1.4\times 10^{-3}\ kg\times 9.8\ m/s^2\times 4\ m\\\\P=0.05488\ J

Finally, it will have kinetic energy. It is given by :

E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 1.4\times 10^{-3}\times (0.9)^2\\\\E=0.000567\ J

The  kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E=0.000567-0.05488=0.054313\ J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.054313 .

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2 years ago
A 11.7-Ω resistor is made from a coil of copper wire whose total mass is 13.5 g . The resistivity of copper is 1.68×10−8Ω⋅m, the
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Answer:

a) d = 7.62 10⁻⁶ m, b)    l = 3.25 10⁴ m

Explanation:

Resistance is expressed by the formula

     R = ρ l / A                       (1)

density is defined by

     density = m / V

the volume of a wire is the cross section by the length

     V = A l

we substitute

     density = m / A l

     A = m / density l

we substitute in 1

     R = ρ l density l / m

     R =ρ density l² / m

     l = √ (R m /ρ density)

let's calculate the cable length

     l = √(11.7  13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))

     l = √(10.56 10⁸)

     l = 3.25 10⁴ m

now we can find the cable diameter with the density equation

      A = m / density l

      A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)

      A = 4,557 10⁻¹¹ m²

the area of ​​the circle is

      A = π r² = π d² / 4

      d = √ (4A /π)

      d = √ (4 4,557 10⁻¹¹/π)

      d = 7.62 10⁻⁶ m

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3 years ago
In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from
Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s     time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s     time =t      

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distance = speed x time

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27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

s = ut + \dfrac{1}{2}at^2

s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2

s = 68.94 m

3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}

3000 = 27.5 t + 68.94 + 13.33 t

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distance = s x t

d = 27.22 x 71.79

d = 1954 m

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distance between A and B is equal to 1.954 Km.

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