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3 years ago

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(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in an Egyptian pyramid, give

n its mass is about 8 ✕ 109 kg and its center of mass is 34.0 m above the surrounding ground?.

1 answer:

Rom4ik [11]3 years ago

3 0

Answer:

$2.66832\times10^{12}$

Or

2,668,320,000

Explanation:

Potential energy is given as the product of mass, acceleration due to gravity and height.

PE=mgh

Where m is the mass of pyramid, g is acceleration due to gravity and h is height at any point.

In reference to the earth, we substitute h for 34.0 m and g with 9.81 m/s2 then taking m as 8000000000 kg we get that

PE=mgh=8000000*9.81*34.0=2,668,320,000 J

In scientific notafion, this is expressed as $2.66832\times10^{12}$

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bazaltina [42]

Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

$R=\frac{25}{100}*74kg=18.5kg$

So the net force uproad on the balloon is

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$P_h=\frac{m}{V}=0.179 kg/m^3$

$P_A=1.29 kg/m^3$

$F_b=F_A-F_H$

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$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

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mezya [45]

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3 years ago

How much work is done to lift a 3N box to a shelf 2m above the ground

Kaylis [27]

When the box IS on the shelf 2m above the ground,
its potential energy is

(weight) x (height) = (3 N) x (2 m) = 6 joules .

THAT's the work you have to do, to lift the box up to there.

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3 years ago

A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir

ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

$E=0.5mV^2$

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1= the cars are separated

2= the cars are togheter

E1=E2

$E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2$

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

$E2=0.5(3)(m)V^2$

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

$0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2$

$V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\$

$V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s$

the speed of the three coupled cars after the first couples with the other two is 2.245m/s

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3 years ago

Read 2 more answers

Two equally charged, 2.098 g spheres are placed with 3.338 cm between their centers. When released, each begins to accelerate at

photoshop1234 [79]

Answer:

$2.64\times 10^{-7} C$

Explanation:

There are two spheres name 1 and 2 and they posses the same charge, which is +q.

And they have equal mass which is 2.098 g.

The distance between these two spheres is, $r=3.338 cm$ .

And the acceleration of each sphere is, $a=269.429 m/s^{2}$ .

Now the coulumbian force experience by 1 sphere due to 2 sphere,

$F_{21} =\frac{q^{2} }{4\pi\epsilon_{0} r^{2} }$ .

And also the newton force will occur due to this force,

$F_{21}=ma$ .

Now equate the above two values of force will get,

$\frac{q^{2} }{4\pi\epsilon_{0} r^{2} } =ma$

Further solve this,

$q^{2}=ma4\pi \epsilon_{0} r^{2}$ .

Substitute all the known variables in above equation,

$q^{2}=(2.098\times 10^{-3} )(269.429)(4(3.14))(8.85\times 10^{-12})(3.338\times 10^{-2})$ .

$q=2.64\times 10^{-7} C$ .

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3 years ago