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3 years ago

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Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g

ravitational potential energy gained by the couch? Estimate g as 9.81 m/s^2. (Answer in Joules)

2 answers:

gavmur [86]3 years ago

4 0

As we know that gravitational potential energy is given by

$U = mgH$

here we have

m = mass = 120 kg

$g = 9.81 m/s^2$

h = height = 8.2 m

now from above formula

$U = 120kg (9.81 m/s^2) (8.2 m)$

$U = 9653.04 J$

so above is the gravitational potential energy of the couch

Natalka [10]3 years ago

3 0

Given data:

height of cough from ground floor (h) = 8.2 m,

mass of the cough (m) = 120 kg,

find gravitational Potential energy (U) =?

acceleration due to gravity (g) = 9.81 m/s²

Gravitational potential energy is defined as "the energy possessed by an object because of its position in the gravitational field". This principle can be applied to objects that are over the earth's surface and gravitational constant assumed to be 9.81 m/s².

Gravitational potential energy also defined as work done in moving the object from infinite to a particular point by the effect of gravity.

Therefore, mathematically

Grvitational Potential energy

U = m.g.h Joules

where,

m= mass of object in kg

g = gravitational constant in m/s² (g=9.81 m/s²)

h = height of the object from ground in m

U = 120 kg × 9.81 m/s²× 8.2 m

= 9653.04 J

<em>The gravitational potential energy is 9653.04 Joules</em>

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Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

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we know that

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solving for

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we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

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3 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

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ma = Eq

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so

put here all value

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so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

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t = 11.80 × $10^{-9}$ s

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time is 23.6 × $10^{-9}$ s

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and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

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Note that I'm reading the given details as

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4 years ago

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Elanso [62]

Answer:

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