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andriy [413]
3 years ago
8

An ice cube (mass = 1.0 kg) slides down an inclined serving tray with an acceleration of 4.0 meters per second. Ignoring frictio

n how much force is pulling the ice cube down the serving tray?
Physics
1 answer:
Agata [3.3K]3 years ago
8 0
Ten newton force is pulling the ice cube towards the tray
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What atom has 5 neutrons and an atomic number of 5?!
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Answer:

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but it has 6 neutron

Explanation:

there is no atom which has 5 neutrons and an atomic number of 5.

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The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t
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The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

PE=\frac{GMm}{r}

Where,

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G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

Then the potential energy at this point would be,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

The we calculate the Potential gravitational energy,

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

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