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3 years ago

A(n) 15.9 kg rock is on the edge of a(n

Physics

1 answer:

Mariulka [41]3 years ago

5 0

The potential energy (P.E) of the rock relative to the base of the cliff is 31,054.926 Joules..

<u>Given the following data:</u>

Mass of rock = 7.5 kg

Height = 0.7 m

Acceleration due to gravity = 9.8 $m/s^2$

To calculate the potential energy (P.E) of the rock relative to the base of the cliff:

<h3>What is potential energy?</h3>

Potential energy (P.E) can be defined as the energy that is possessed by an object due to its position (height) above planet Earth.

Mathematically, potential energy (P.E) is given by this formula;

$P.E = mgh$

<u>Where:</u>

P.E is the potential energy.

m is the mass of an object.

g is the acceleration due to gravity.

h is the height of an object.

Substituting the given parameters into the formula, we have;

$P.E = 15.9 \times 9.8 \times 199.3$

P.E = 31,054.926 Joules.

Read more on potential energy here: brainly.com/question/8664733

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A 1kw electric heater is switched on for ten minutes. how much heat does it produce

Ainat [17]

H= P × t

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10 min = 600s

H= 1000×600=600,000J

=> 143.40kcal

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3 years ago

A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o

maxonik [38]

For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
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3 years ago

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2 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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3 years ago

Read 3 more answers

A water bottle is dropped from a plane and accelerates because of gravity. What distance did it fall during 3 seconds?

exis [7]

It’d fall 29.4m or 96.46ft

Explanation:
Uhh since gravity is 9.8m/s then in three seconds it’d drop 29.4m or 96.46ft
That is assuming there isn’t a lot of wind resistance, but if you take that into account, then it’d probably be somewhere around 25m since the water bottle is going to be heavier than the wind resistance, and since we don’t know the weight of the water bottle it can’t really be calculated.

Hope this helps!

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2 years ago