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2 years ago

Explain the forces that keep a rock balanced on a tiny pedestal. Please help me with this!

1 answer:

8 0

Normal force for the rock because that makes an object stable at its position.
static friction because micro-welts hold its particle on its position so it doesn't change in position by a potential energy. Gravity makes it stay on the ground because its force attraction between an object and the earth.
Hope this helps <span />

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engineers who design battery operated devices suck as sell phones and MP3 players try to make them as efficient as possible. An

ICE Princess25 [194]

If the 5,500 J of sound and light is the ONLY useful output
from the phone, then the phone's efficiency is

(5,500J / 10,000J) = 0.55 = 55% .

But the test engineer forgot one little minor almost insignificant detail.
As a test engineer myself, I'd say that he needs to turn in his laptop
and soldering iron, and think about changing his career to a job for
which he may be better suited, like 8 hours a day in a highway toll-booth.

What about that little radio transmitter and receiver inside the phone,
that maintain digital RF communication with the nearest cell tower ?
Without that microscopic radio transceiver ... plus 30 or 40 apps
that are always running unless you shut them off ... the device in your
pocket is essentially a flat rock with one side that sometimes glows.

8 0

3 years ago

one car accelerates at half the rate of another how much longer does it take the first car to travel a quarter mile

Orlov [11]

Answer:

t=1/4v1

Explanation:

Given data

Car one

Speed =v1

Time =t

Distance =1/4 mile

Given data

Car two

Speed =v1/2

Time =t

Distance =d

Speed =distance/time

v1=1/4/t

v1t=1/4

t=1/4*1/v1

t=1/4v1 seconds

7 0

2 years ago

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a

harina [27]

Answer:

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

Em_f = U = - G Mm / R₂

energy is conserved

Em₀ = Em_f

½ m v_c² - G Mm / R = - G Mm / R₂

v_c² = 2 G M (1 /R - 1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$ )

we make the product and keep the terms linear

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

5 0

2 years ago

What happened to the weight of an object when it is taken from Earth to the Moon? why?<br>

Sholpan [36]

Answer:

the weight of the object decreases when it is taken from the Earth to the Moon

Explanation:

The weight of an object is defined as the product of the mass of the object with the acceleration due to gravity of the Planet.

$W =mg$

where,

W = weight of the object

m = mass of the object

g = acceleration due to gravity on the planet

The mass of an object remains constant everywhere in the universe. Therefore, the weight is directly proportional to the value of acceleration due to gravity.

The value of acceleration due to gravity on the Moon is lesser than its value on the Earth.

<u>Hence, the weight of the object decreases when it is taken from the Earth to the Moon </u>

6 0

3 years ago

What kind of device can you use to to separate visible light into its different colors

juin [17]

A prism will separate white light into a rainbow of light

5 0

3 years ago